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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using static System.Math;
 
public class Test
{
	public static void Main()
	{
		// my code goes here
		
		var hoi = Console.ReadLine().Split().Select(nyan => int.Parse(nyan)).ToArray(); //この呪文覚えたい
		int n = hoi[0];
		int d = hoi[1];
		
		
		//ここから2次元配列を召喚
		double[,] x = new double[n,d];
		for(int  i=0; i!=n; i++){
			var gyou = Console.ReadLine().Split().Select(nyaa => double.Parse(nyaa)).ToArray();
			for(int j=0; j!=d; j++){
				x[i,j] = gyou[j];
			}
		}
		
		//入力完了、あとは計算
		
		int counter = 0;
		double distance = 0;
		double dis2zyou = 0;
		for(int k=0; k!=n; k++){
			for(int l=k+1; l<n; l++){
				for(int m=0; m!=d; m++){dis2zyou += Pow( x[k,m] - x[l,m] ,2);}
				distance = Sqrt(dis2zyou);
				if(distance - Floor(distance) == 0){counter++;} //GOOD DISTANCE!!
				distance = 0;
				dis2zyou = 0;
			}
		}
		Console.WriteLine(counter);
	}
}