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: Wrong Answer)

#include <iostream>
#include <cmath>
#include <vector>

using namespace std;

#define PROD

int main()
{
    int N, D;

#ifdef PROD
    cin >> N >> D;

    vector<vector<int>> X(N, vector<int>(D));
    for(int ii = 0 ; ii < N ; ii++) {
        for(int ij = 0 ; ij < D ; ij++) {
            cin >> X[ii][ij];
        }
    }
#else
    N = 3;
    D = 2;

    vector<vector<int>> X(N, vector<int>(D));
    X[0][0] = 1;
    X[0][1] = 2;
    X[1][0] = 5;
    X[1][1] = 5;
    X[2][0] = -2;
    X[2][1] = 8;
#endif

    auto ans = 0;

    for(int i = 0 ; i < N - 1 ; i++) {
        for(int j = i + 1; j < N ; j++) {
            int sum = 0;
            for(int k = 0 ; k < D ; k++) {
                auto diff = abs(X[i][k] - X[j][k]);
                sum += diff * diff;
            }

            bool f = false;
            for(int k = 0; k < sum; k++) {
                if(k * k == sum) {
                    f = true;
                }
            }

            if (f) ans++;
        }
    }

    cout << ans << endl;
    return 0;
}