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 pb push_back
#define mp make_pair
#define se second
#define fi first
#define ll long long int
#define INF INT_MAX
#define mod
#define chutiyaap ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
using namespace std;
bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b){return (a.second < b.second);}
ll gcd(ll a, ll b){if(a==0)return b;return gcd(b % a, a);}
int main()
{
	chutiyaap;
	ll n,d;cin>>n>>d;
	ll a[n][d];for(ll i=0;i<n;i++)for(ll j=0;j<d;j++)cin>>a[i][j];
	ll res=0;
	for(ll i=0;i<n;i++)
	{
		for(ll j=i+1;j<n;j++)
		{
			double dis=0;
			for(ll k=0;k<d;k++)
			{
				dis+=pow((abs(a[i][k]-a[j][k])),2);
			}
			dis=sqrt(dis);
			ll dist=dis;
			if(dist==dis)res++;
		}
	}
	cout<<res;
}