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;

#define pf(n) cout << n << endl
#define ps() cout <<" ";
#define sf(n) cin >> n
#define in(i,n) for (int i = 0; i < n; i++)
#define vi vector<int>
#define vp vector<pair<int, int>>
#define graph vector< vector<int> >
#define vs vector<string>

int x;
int main() {

  //ios::sync_with_stdio(false);
  //cin.tie(0);
  int n;
  int d;
  sf(n);sf(d);
  float e;
  vector<vector<float> > a(n);
  in(i,n)
  {
      in(j,d)
      {
          sf(e);
          a[i].push_back(e);
      }
  }
  int count=0;
  for(int i=0;i<n;i++)
  {
      for(int j=i+1;j<n;j++)
      {
          float sum=0;
          for(int k=0;k<d;k++)
          {
              sum+=((a[i][k]-a[j][k])*(a[i][k]-a[j][k]));
          }
          sum=sqrt(sum);
          //pf(sum);
          if(sum==floor(sum))
            count++;
      }
  }
  pf(count);
}