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 For(i,a,b) for(int i=a;i<=b;i++)
#define F first
#define S second
#define all(x) x.begin(),x.end()
#define sz(x) ((int)x.size())
#define int ll
#define MOD (ll)(1e9+7)
using namespace std;
using ll=long long;
using pii=pair<int,int>;

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int n,d; cin>>n>>d;
	vector<vector<int>> v(n);
	for(auto &i:v)
	{
		i.resize(d);
		for(auto &j:i) cin>>j;
	}
	int cnt=0;
	For(i,0,n-2)
	{
		For(j,i+1,n-1)
		{
			int dis=0;
			For(k,0,d-1) dis+=pow(v[i][k]-v[j][k],2);
			if(pow(floor(sqrt(dis)),2)==dis) cnt++;
		}
	}
	cout<<cnt<<"\n";
	return 0;
}