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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define speed ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define ub(a,x) upper_bound(a.begin(),a.end(),x)-a.begin()
#define lb(a,x) lower_bound(a.begin(),a.end(),x)-a.begin()
#define ALL(v)   v.begin(),v.end()
int main()
{
	speed
	ll n ,d;
	cin >> n >> d;
	ll vec[n][d];
	for(int x =0; x<n; x++)
	{
		for(int y = 0; y<d; y++)cin >> vec[x][y];
	}
int ans =0 ;
	for(int x =0; x<n; x++)
	{
		for(int y =x+1; y<n; y++)
		{
			ll s = 0;bool r = true;

			for(int z =0; z<d; z++)
			{
				if(vec[x][z] == vec[y][z])r = false;
				s += pow(vec[x][z]- vec[y][z], 2LL);
			}
			if(r && sqrt(s ) * sqrt(s) == s)ans++;
		}
	}
	cout << ans;
	return 0;
}