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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define rep(i,n) for(int i=0;i<n;i++)
int gcd(int a, int b){
    int r=1;
    while(r!=0){
        r=a%b;
        a=b;
        b=r;
    }
    return a;
}
typedef long long ll;
void revStr(char* str){
    int size = strlen(str);
    int i,j;
    char tmp = {0};
    for(i = 0, j = size - 1; i < size / 2; i++, j--){
        tmp = str[i];
        str[i] = str[j];
        str[j] = tmp;
    }return;
}
typedef struct{
    int a,b;
}data;

//昇順
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}
//構造体降順
int comp( const void *c1, const void *c2 )
{
  data test1 = *(data *)c1;
  data test2 = *(data *)c2;
  int tmp1 = test1.b;   /* b を基準とする */
  int tmp2 = test2.b;
  return tmp2 - tmp1;
}


int kai(int n){
    if(n==0){
        return 1;
    }
    int sum=1;
    for(int i=1;i<=n;i++){
        sum*=i;
    }
    return sum;
}

int main(){
    int N,D;
    int cnt=0;
    scanf("%d%d",&N,&D);
    int ten[N][D];
    rep(i,N){
        rep(j,D){
            scanf("%d",&ten[i][j]);
        }
    }
    rep(i,N-1){
        for(int j=i+1;j<N;j++){
            int temp=0;
            for(int k=0;k<D;k++){
                temp+=(ten[i][k]-ten[j][k])*(ten[i][k]-ten[j][k]);
            }
            for(int m=1;m<=16000;m++){
                if(temp==m*m){
                    cnt++;
                }
            }
        }
    }
    printf("%d",cnt);
}