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;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define len(v) ll(v.size())

template <class T>
void cout_vec(const vector<T> &vec){
  for(auto itr:vec) cout<<itr<<' ';
  cout<<'\n';
}

typedef pair<ll,ll> P;
const ll mod=1e9+7;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,d;
    cin>>n>>d;
    int x[20][20];
    FOR(i,1,n+1)FOR(j,1,d+1) cin>>x[i][j];
    int ans=0;
    FOR(i,1,n+1)FOR(j,1,n+1){
      if(i>=j) continue;
      int now=0;
      FOR(k,1,d+1){
        now+=(x[i][k]-x[j][k])*(x[i][k]-x[j][k]);
        //cout<<now<<endl;
      }
      rep(i,100000){
        if(i*i==now){
          ans++;
          //cout<<now<<endl;
        }
      }
    }
    cout<<ans<<endl;
}