Minimize distance between two lists











up vote
5
down vote

favorite
1












Writing:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



On the other hand, if I write:



k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.



Question: How can I determine the best value of k to get the smallest possible gap?





Writing:



h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



Question 2: is it possible to determine the pair of values h, k that minimize the gap?










share|improve this question
























  • This is a related question: How to find the distance of two lists?
    – Artes
    Nov 29 at 16:03















up vote
5
down vote

favorite
1












Writing:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



On the other hand, if I write:



k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.



Question: How can I determine the best value of k to get the smallest possible gap?





Writing:



h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



Question 2: is it possible to determine the pair of values h, k that minimize the gap?










share|improve this question
























  • This is a related question: How to find the distance of two lists?
    – Artes
    Nov 29 at 16:03













up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Writing:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



On the other hand, if I write:



k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.



Question: How can I determine the best value of k to get the smallest possible gap?





Writing:



h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



Question 2: is it possible to determine the pair of values h, k that minimize the gap?










share|improve this question















Writing:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



On the other hand, if I write:



k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.



Question: How can I determine the best value of k to get the smallest possible gap?





Writing:



h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


I get:



enter image description here



Question 2: is it possible to determine the pair of values h, k that minimize the gap?







mathematical-optimization charts






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 28 at 21:27

























asked Nov 28 at 20:19









TeM

1,777619




1,777619












  • This is a related question: How to find the distance of two lists?
    – Artes
    Nov 29 at 16:03


















  • This is a related question: How to find the distance of two lists?
    – Artes
    Nov 29 at 16:03
















This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03




This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03










2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










Update: Using two parameters:



lmf2 = LinearModelFit[data, t, t];
Normal@lmf2



1.76563 + 0.546875 t




lmf2["BestFitParameters"]



{1.76563, 0.546875}




Fit[data, {1, t}, t]



1.76563 + 0.546875 t




ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]



{43.3594, {h -> 1.76562, k -> 0.546875}}




N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]



{1.76563, 0.546875}




Original answer:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];


You can use LinearModelFit or Fit or NMinimize or LeastSquares to get the value of k that minimizes the sum of squared distances between expectedresults and k achievedresults:



lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]

Normal@lmf



0.828025 t




Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]



0.828025 #1




Fit[data, {t}, t]



0.828025 t




ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]



{49.7134, {k -> 0.828025}}




N@LeastSquares[Thread[{achievedresults}], expectedresults]



{0.828025}




k = lmf["BestFitParameters"][[1]]



0.828025




p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


enter image description here



BarChart[Transpose@{expectedresults, achievedresults,  k achievedresults}, 
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]


enter image description here






share|improve this answer























  • For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
    – David G. Stork
    Nov 28 at 20:42










  • @TeM, please see the update.
    – kglr
    Nov 28 at 21:47










  • Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
    – TeM
    Nov 28 at 21:48






  • 1




    @TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
    – kglr
    Nov 28 at 21:54


















up vote
7
down vote













{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults



{35/64, 113/64}







share|improve this answer



















  • 2




    Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
    – MeMyselfI
    Nov 28 at 20:56






  • 1




    @MeMyselfI Nice! Even better
    – Chris
    Nov 28 at 21:05










  • Really great!!!
    – TeM
    Nov 28 at 21:50











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Update: Using two parameters:



lmf2 = LinearModelFit[data, t, t];
Normal@lmf2



1.76563 + 0.546875 t




lmf2["BestFitParameters"]



{1.76563, 0.546875}




Fit[data, {1, t}, t]



1.76563 + 0.546875 t




ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]



{43.3594, {h -> 1.76562, k -> 0.546875}}




N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]



{1.76563, 0.546875}




Original answer:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];


You can use LinearModelFit or Fit or NMinimize or LeastSquares to get the value of k that minimizes the sum of squared distances between expectedresults and k achievedresults:



lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]

Normal@lmf



0.828025 t




Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]



0.828025 #1




Fit[data, {t}, t]



0.828025 t




ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]



{49.7134, {k -> 0.828025}}




N@LeastSquares[Thread[{achievedresults}], expectedresults]



{0.828025}




k = lmf["BestFitParameters"][[1]]



0.828025




p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


enter image description here



BarChart[Transpose@{expectedresults, achievedresults,  k achievedresults}, 
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]


enter image description here






share|improve this answer























  • For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
    – David G. Stork
    Nov 28 at 20:42










  • @TeM, please see the update.
    – kglr
    Nov 28 at 21:47










  • Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
    – TeM
    Nov 28 at 21:48






  • 1




    @TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
    – kglr
    Nov 28 at 21:54















up vote
4
down vote



accepted










Update: Using two parameters:



lmf2 = LinearModelFit[data, t, t];
Normal@lmf2



1.76563 + 0.546875 t




lmf2["BestFitParameters"]



{1.76563, 0.546875}




Fit[data, {1, t}, t]



1.76563 + 0.546875 t




ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]



{43.3594, {h -> 1.76562, k -> 0.546875}}




N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]



{1.76563, 0.546875}




Original answer:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];


You can use LinearModelFit or Fit or NMinimize or LeastSquares to get the value of k that minimizes the sum of squared distances between expectedresults and k achievedresults:



lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]

Normal@lmf



0.828025 t




Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]



0.828025 #1




Fit[data, {t}, t]



0.828025 t




ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]



{49.7134, {k -> 0.828025}}




N@LeastSquares[Thread[{achievedresults}], expectedresults]



{0.828025}




k = lmf["BestFitParameters"][[1]]



0.828025




p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


enter image description here



BarChart[Transpose@{expectedresults, achievedresults,  k achievedresults}, 
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]


enter image description here






share|improve this answer























  • For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
    – David G. Stork
    Nov 28 at 20:42










  • @TeM, please see the update.
    – kglr
    Nov 28 at 21:47










  • Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
    – TeM
    Nov 28 at 21:48






  • 1




    @TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
    – kglr
    Nov 28 at 21:54













up vote
4
down vote



accepted







up vote
4
down vote



accepted






Update: Using two parameters:



lmf2 = LinearModelFit[data, t, t];
Normal@lmf2



1.76563 + 0.546875 t




lmf2["BestFitParameters"]



{1.76563, 0.546875}




Fit[data, {1, t}, t]



1.76563 + 0.546875 t




ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]



{43.3594, {h -> 1.76562, k -> 0.546875}}




N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]



{1.76563, 0.546875}




Original answer:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];


You can use LinearModelFit or Fit or NMinimize or LeastSquares to get the value of k that minimizes the sum of squared distances between expectedresults and k achievedresults:



lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]

Normal@lmf



0.828025 t




Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]



0.828025 #1




Fit[data, {t}, t]



0.828025 t




ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]



{49.7134, {k -> 0.828025}}




N@LeastSquares[Thread[{achievedresults}], expectedresults]



{0.828025}




k = lmf["BestFitParameters"][[1]]



0.828025




p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


enter image description here



BarChart[Transpose@{expectedresults, achievedresults,  k achievedresults}, 
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]


enter image description here






share|improve this answer














Update: Using two parameters:



lmf2 = LinearModelFit[data, t, t];
Normal@lmf2



1.76563 + 0.546875 t




lmf2["BestFitParameters"]



{1.76563, 0.546875}




Fit[data, {1, t}, t]



1.76563 + 0.546875 t




ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]



{43.3594, {h -> 1.76562, k -> 0.546875}}




N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]



{1.76563, 0.546875}




Original answer:



expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];


You can use LinearModelFit or Fit or NMinimize or LeastSquares to get the value of k that minimizes the sum of squared distances between expectedresults and k achievedresults:



lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]

Normal@lmf



0.828025 t




Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]



0.828025 #1




Fit[data, {t}, t]



0.828025 t




ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]



{49.7134, {k -> 0.828025}}




N@LeastSquares[Thread[{achievedresults}], expectedresults]



{0.828025}




k = lmf["BestFitParameters"][[1]]



0.828025




p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]


enter image description here



BarChart[Transpose@{expectedresults, achievedresults,  k achievedresults}, 
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 28 at 21:52

























answered Nov 28 at 20:29









kglr

175k9197402




175k9197402












  • For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
    – David G. Stork
    Nov 28 at 20:42










  • @TeM, please see the update.
    – kglr
    Nov 28 at 21:47










  • Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
    – TeM
    Nov 28 at 21:48






  • 1




    @TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
    – kglr
    Nov 28 at 21:54


















  • For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
    – David G. Stork
    Nov 28 at 20:42










  • @TeM, please see the update.
    – kglr
    Nov 28 at 21:47










  • Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
    – TeM
    Nov 28 at 21:48






  • 1




    @TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
    – kglr
    Nov 28 at 21:54
















For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42




For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42












@TeM, please see the update.
– kglr
Nov 28 at 21:47




@TeM, please see the update.
– kglr
Nov 28 at 21:47












Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48




Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48




1




1




@TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
– kglr
Nov 28 at 21:54




@TeM, If you compare the NMinimize result adding the intercept parameter improves the squared loss from 49.7134 to 43.3594.
– kglr
Nov 28 at 21:54










up vote
7
down vote













{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults



{35/64, 113/64}







share|improve this answer



















  • 2




    Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
    – MeMyselfI
    Nov 28 at 20:56






  • 1




    @MeMyselfI Nice! Even better
    – Chris
    Nov 28 at 21:05










  • Really great!!!
    – TeM
    Nov 28 at 21:50















up vote
7
down vote













{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults



{35/64, 113/64}







share|improve this answer



















  • 2




    Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
    – MeMyselfI
    Nov 28 at 20:56






  • 1




    @MeMyselfI Nice! Even better
    – Chris
    Nov 28 at 21:05










  • Really great!!!
    – TeM
    Nov 28 at 21:50













up vote
7
down vote










up vote
7
down vote









{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults



{35/64, 113/64}







share|improve this answer














{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults



{35/64, 113/64}








share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 28 at 21:45

























answered Nov 28 at 20:36









Chris

54116




54116








  • 2




    Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
    – MeMyselfI
    Nov 28 at 20:56






  • 1




    @MeMyselfI Nice! Even better
    – Chris
    Nov 28 at 21:05










  • Really great!!!
    – TeM
    Nov 28 at 21:50














  • 2




    Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
    – MeMyselfI
    Nov 28 at 20:56






  • 1




    @MeMyselfI Nice! Even better
    – Chris
    Nov 28 at 21:05










  • Really great!!!
    – TeM
    Nov 28 at 21:50








2




2




Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
– MeMyselfI
Nov 28 at 20:56




Why do you add a zero column? I think PseudoInverse[Transpose[{achievedresults}]].expectedresults will do
– MeMyselfI
Nov 28 at 20:56




1




1




@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05




@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05












Really great!!!
– TeM
Nov 28 at 21:50




Really great!!!
– TeM
Nov 28 at 21:50


















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