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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5+10 , OO = 0x3f3f3f3f;

#define clr(arr,val) memset(arr , val , sizeof (arr))
#define loop(i,n) for(int i = 0;i < int(n);i++)
#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)
#define range(i,a,b) for(int i = int(a);i <= int(b);i++)
#define ALL(v) ((v).begin()) , ((v).end())
#define SZ(v)  ((int)((v).size()))
typedef pair <int,int> pii ;
typedef vector < pii > vii ;
typedef vector < int > vi;
int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction
//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction
//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);


int point[15][15];
int n ,d;


int main()
{   

      ios_base::sync_with_stdio(0);
      cin.tie(0);cout.tie(0);

      cin >> n >> d;
      loop(i, n){
            loop(j, d)
                  cin >> point[i][j];
      }      
      int cnt = 0;  // dis = 0;
      loop(i , n){
            range(j , i+1 , n-1){
                  int dis = 0;
                  loop(dimen, d){
                        dis += ((point[i][dimen]-point[j][dimen])*(point[i][dimen]-point[j][dimen]));
                  }
                  double ans = sqrt(dis);
                  int rkm = ans;
                  if(rkm-ans == 0) cnt++;
            }
      }
      cout << cnt << "\n";

      return 0;
}