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>
#define rep(i, n) for(int i=0 ;i <(n); ++i)
typedef long long ll;
using namespace std;

int pow2Sum(vector<int> a, vector<int>b) {
  int res = 0;
  int size = a.size();
  rep(i, size) res += (a[i] - b[i]) * (a[i] - b[i]);
  return res;
}

int main() {
  int n, d;
  cin >> n >> d;
  vector<vector<int>> a(n, vector<int>(d));
  rep(i, n) {
    rep(j, d) {
      cin >> a[i][j];
    }
  } 
  int c = 0;

  // int test = pow2Sum(a[0], a[1]);
  // double sqTest = sqrt(test);

  // cout << test << endl; 
  // cout << sqTest - floor(sqTest) << endl; 

  rep(i, n) {
    rep(j, n) {
      if(i != j) {
        double num = sqrt(pow2Sum(a[i], a[j]));
        if(num - floor(num) == 0) ++c;
      }
    }
  }

  cout << c/2 << endl;
  return 0;
}