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 main(void){
    int N;
    int D;

    scanf("%d",&N);
    scanf("%d",&D);

    int zahyo[N][D];

    int i;
    int j;
    for(i=0;i<N;i++){
        for(j=0;j<D;j++){      
            scanf("%d",&zahyo[i][j]);
        }
    }

    double keisankekka=0;
    double heihouwa=0;
    int jud=0;

    int counter=0;
    
    for(i=0;i<N;i++){
        j=0;
        for(j=0;j<D;j++){
            if(i+1>D){
                break;
            }
            keisankekka +=(zahyo[i][j]-zahyo[i+1][j])*(zahyo[i][j]-zahyo[i+1][j]);
        }
        
        heihouwa=sqrt(keisankekka);
        jud=pow((float)keisankekka,0.5);
        if(heihouwa==(double)jud){
            counter++;
        }
    }


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