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>
using namespace std;
using ll=long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
#define all(x) x.begin(),x.end()
#define rep(i,j,n) for (int i = j; i < (int)(n); i++)
#define _GLIBCXX_DEBUG
#define MOD 1000000007
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//name[i][j]-> vvl name(i,vl(j))
//map<int,int>mp
//余りを切り上げる =>  (a + b - 1) / b 

signed main(){
	int n,d; cin>>n>>d;
	vvi v(n,vi(d));
	rep(i,0,n){
		rep(j,0,d){
			cin>>v[i][j];
		}
	}
	int ans=0;
	rep(i,0,n-1){
		rep(j,i+1,n){
			int l=0;
			rep(k,0,d){
				int p=(v[i][k]-v[j][k]);
				p*=p;
				l+=p;
			}
			//cout<<"l:"<<l<<endl;
			int y=sqrt(l);
			if(y*y==l) ans++;
		}
	}
	cout<<ans<<endl;
	
	return 0;
}