Problem:
Score : 200 points

Problem Statement

There are N points in a D-dimensional space.

The coordinates of the i-th point are (X_{i1}, X_{i2}, ..., X_{iD}).

The distance between two points with coordinates (y_1, y_2, ..., y_D) and (z_1, z_2, ..., z_D) is \sqrt{(y_1 - z_1)^2 + (y_2 - z_2)^2 + ... + (y_D - z_D)^2}.

How many pairs (i, j) (i < j) are there such that the distance between the i-th point and the j-th point is an integer?

Constraints

All values in input are integers.

2 \leq N \leq 10

1 \leq D \leq 10

-20 \leq X_{ij} \leq 20

No two given points have the same coordinates. That is, if i \neq j, there exists k such that X_{ik} \neq X_{jk}.

Input

Input is given from Standard Input in the following format:

N D
X_{11} X_{12} ... X_{1D}
X_{21} X_{22} ... X_{2D}
\vdots
X_{N1} X_{N2} ... X_{ND}

Output

Print the number of pairs (i, j) (i < j) such that the distance between the i-th point and the j-th point is an integer.

Sample Input 1

3 2
1 2
5 5
-2 8

Sample Output 1

1

The number of pairs with an integer distance is one, as follows:

The distance between the first point and the second point is \sqrt{|1-5|^2 + |2-5|^2} = 5, which is an integer.

The distance between the second point and the third point is \sqrt{|5-(-2)|^2 + |5-8|^2} = \sqrt{58}, which is not an integer.

The distance between the third point and the first point is \sqrt{|-2-1|^2+|8-2|^2} = 3\sqrt{5}, which is not an integer.

Sample Input 2

3 4
-3 7 8 2
-12 1 10 2
-2 8 9 3

Sample Output 2

2

Sample Input 3

5 1
1
2
3
4
5

Sample Output 3

10

Solution in Ruby (2.3.3) (judge verdict: Accepted)

# Your code here!
input=gets.split(" ")
N=input[0].to_i
D=input[1].to_i
X=[]
for i in 0..N-1
 X[i]=gets.split(" ")
end
for i in 0..N-1
 for j in 0..D-1
  X[i][j]=X[i][j].to_i
 end
end

comb=N*(N-1)/2


#点0と点２の距離を測る実験
sig=[]
sum=[]
for i in 0..comb-1
 sum[i]=0
end

pre_sum = Array.new(N).map{Array.new(N,0)}

for h in 0..N-1
 for i in 0..N-1
  pre_sum[h][i]=0
 end
end
 L=N
 for h in 0..N-1
 for i in 0..L-1
  for j in 0..D-1#次元数ループ
   sig[j]=(X[h][j]-X[i][j])*(X[h][j]-X[i][j])
   pre_sum[h][i]=pre_sum[h][i]+sig[j]
  end
 end
 end
 pre_count=0
 for h in 0..N-1
 for i in 0..L-1
 if Math.sqrt(pre_sum[h][i])==Math.sqrt(pre_sum[h][i]).to_i && Math.sqrt(pre_sum[h][i])!=0
     pre_count=pre_count+1
 end
 end
 end
 count=pre_count/2
 puts(count)