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 sz(a) (int)(a).size()
     
    using namespace std;
     
    template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
    template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
    const int INF = 2e5+2;
    const long long hell = 998244353ll;
    
    double dis(vector<int>a,vector<int> b){
    	double ans=0;
    	for(int i=0; i<sz(a); i++){
    		ans += (a[i]-b[i])*(a[i]-b[i]);
    	}
    	return sqrt(ans);
    }

    void solve(){
          int n,d;
          cin>>n>>d;
          vector<vector<int> > arr(n,vector<int> (d));
          for(int i=0; i<n; i++){
          	for(int j=0; j<d; j++){
          		cin>>arr[i][j];
          	}
          }
          int cnt=0;
          double ans=0;
          for(int i=0; i<n; i++){
          	for(int j=0; j<n; j++){
          		ans = dis(arr[i],arr[j]);
          		if(i!=j && ceil(ans)==floor(ans)){
          			cnt++;
          		}
          	}
          }

          cout<<cnt/2<<endl;

    }

    int main(){
    	
    	ios_base::sync_with_stdio(0);cin.tie(0);
    	
    	int q=1;
    #ifdef LOCAL
      	cin>>q;
#endif
	while(q--){
	 solve();
	}

    #ifdef LOCAL
        cout << "Time elapsed: " << 1000.0 * clock() / CLOCKS_PER_SEC << " ms.\n";
    #endif	
    	return 0;
    }