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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
int D,N;
/*int dis(int y[N][D],int z[N][D]){
  int i,k=0;
  double x;
  for(i=0;i<D;i++)
    x+=pow(y[][i]-z[][i],2);
  x = sqrt(x);
  if(x-(int)x == 0.0)
    return 1;
  else
    return -1;
}*/

int main(){
  int i,j,k=0,N,ans=0;
  double x;
  scanf("%d %d",&N,&D);
  double y[N][D];
  for(i=0;i<N;i++){
    for(j=0;j<D;j++)
      scanf("%lf",&y[i][j]);
  }
  
  for(i=0;i<N;i++){
    for(j=i+1;j<N;j++){
      x=0;
      for(k=0;k<D;k++){
        x+=pow(y[i][k]-y[j][k],2);
      }
      x=sqrt(x);
      //printf("%lf\n",x);
      if(ceil(x) == floor(x))
        ans++;
    }
  }
  
  printf("%d\n",ans);
  return 0;
}