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 isPower(int a, int d);

int main(void){
  int i,j,k;
  int n,d;
  int sum=0;
  int count=0;

  scanf("%d %d",&n, &d);
  int x[n][d];

  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+1; j<n; j++){
      //x[i][]とx[i+1][],x[i+2][],...,x[n-1][]の比較
      sum=0;
      for(k=0; k<d; k++){
        //x[i][]とx[j][]の比較
        //n乗根にする前の値
        sum+=pow(x[i][k]-x[j][k],2);
        //sumがd乗根だったらcount++
        // printf("i=%d j=%d k=%d\n", i,j,k);
        // printf("x[i][k]-x[j][k]=%d\n", x[i][k]-x[j][k]);
      }
      // printf("sum=%d\n", sum);
      if(isPower(sum, 2)) count++;
    }
  }

  printf("%d\n", count);
  return 0;
}

int isPower(int a, int d){
  int r=1;
  int num;

  while(1){
    num=pow(r,d);
    if(a<num) return 0;
    if(a==num) return 1;
    r++;
  }
}