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 <stdlib.h>
#include <math.h>

int ZettaichiNizyou(int a, int b){
        int N = abs(a-b);
        int Ns = N*N;
        return Ns;
    }

int Check(int a){
    double x = (double)a;
    double y = sqrt(x);
    if(ceil(y)==floor(y)){
        return 1;
    }
    return 0;
}

int main(void){
    int N,D;

    scanf("%d%d",&N,&D);
    
    int a[N][D];
    int n;
    int d;

    for(n = 0;n < N;n++){
        for(d=0;d < D; d++){
            scanf("%d",&a[n][d]);
        }
    }

    int n2,d2;
    int ZN =0;
    int checkcount = 0;
    for(n2=0; n2 < N; n2++){
        int counter = 1;
        while(n2+counter < N){
            for(d2=0; d2 < D; d2++){
                ZN += ZettaichiNizyou(a[n2][d2],a[n2+counter][d2]);
            }
            if(Check(ZN) == 1){
                checkcount++;
            }
            ZN=0;
            counter++;
        }
    }
    printf("%d\n",checkcount);
}