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>
#include<stdlib.h>
int main(){
    int N,D;
    scanf("%d %d",&N,&D);
    int i,j;
    int X[10][10];
    //X = (int **)malloc(N*sizeof(int));
    //for(i=0;i<N;i++){
      //  X[i]= (int *)malloc(D*sizeof(int));
    //}
    for(i=0;i<N;i++){
        for(j=0;j<D;j++){
            scanf("%d",&X[i][j]);
        }
   }
    double d;
    double sum;
    i=0,j=1;
    int count=0;
    int t;
        while(i!=N-1){
        sum = 0;
        //printf("%lf\n",sum);
        for(t=0;t<D;t++){
        //printf("hikizan=%d\n",X[i][t]-X[j][t]);
        sum=sum+pow((X[i][t]-X[j][t]),2);
        //printf("sum=%lf",sum);
        }
        //printf("%lf\n",sum);
        d=sqrt(sum);
        if(ceil(d)==floor(d)){
            count++;
        }
        j++;
        if(j==N){
            i++;
            j=i+1;
        }
        }
    printf("%d\n",count);   
    //for(i=0;i<N;i++){
     //   free(X[i]);
   // }
    return 0;
}