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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
namespace Sample
{
  class Sample
  {
    static void Main(string[] args)
    {
    string[] str=Console.ReadLine().Split(' '); 
      int N=int.Parse(str[0]);
      int D=int.Parse(str[1]);
      int[][] zahyo =new int[N][];
      for(int i=0;i<N;i++){
       string[] str2=Console.ReadLine().Split(' ');
       zahyo[i]=str2.Select(int.Parse).ToArray();
      }
      int ans=0;
      for(int i=0;i<N;i++){
       for(int j=i+1;j<N;j++){     
       int DisSq=0;
       int dis=(int)Math.Sqrt(DisSq);
       for(int k=0;k<D;k++){
         DisSq=(int)Math.Pow(zahyo[i][k]-zahyo[j][k],2);
       }
         if(Math.Pow(dis,2)==DisSq){
           ans++;
         }
         
       }
       
      }
      
       Console.WriteLine(ans);
    }
    
  }

}