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: Wrong Answer)

#include<stdio.h>
#include<math.h>
int distance(int a,int b){
  return (a-b)*(a-b);
}
int main (void)
{
  int n,d,i,j,k;
  int cnt=0;
  scanf("%d%d",&n,&d);
  int x[n][d];
  long long int dis[n][n+1];
  for(i=0;i<n;i++){
    for(j=0;j<d;j++){
      scanf("%d",&x[i][j]);
    }
  }
  for(i=0;i<n-1;i++){
    for(j=i;j<n;j++){
      dis[i][j]=0;
    }
  }


  for(i=0;i<n-1;i++){
    for(j=i+1;j<n;j++){
      for(k=0;k<d;k++){
        dis[i][j]+=distance(x[i][k],x[j][k]);
      }
    }
  }
  /*
  *for(i=0;i<n;i++){
  *  for(j=0;j<d;j++){
  *    printf("x[%d][%d]= %3d   ",i,j,x[i][j]);
  *  }
  *  putchar('\n');
  *}
  */
  for(i=0;i<n-1;i++){
    for(j=i+1;j<n;j++){
      //printf("dis[%d][%d]=%4d ",i,j,dis[i][j]);
      for(k=1;k<41;k++){
        if(sqrt(dis[i][j])==k){
          cnt++;
          break;
        }
      }
    }
    putchar('\n');
  }
  printf("%d",cnt);
  return 0;
}