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.Collections.Generic;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            long[] iArray = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).ToArray();
            long N = iArray[0];
            long D = iArray[1];
            long[,] point = new long[N, D];
            int cnt = 0;
            for (int p = 0; p < N; p++)
            {
                long[] tmpArray = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).ToArray();
                for(int t = 0; t < D; t++)
                {
                    point[p, t] = tmpArray[t];
                }
            }
            
            for (int i = 0;i < N; i++)
            {
                for (int j = i + 1; j < N; j++)
                {
                    long sum = 0;
                    for (int z = 0; z < D; z++)
                    {
                         sum += (long)Math.Pow(point[i, z] - point[j, z], 2);
                    }
                  	long b = (long)Math.Sqrt(sum);
                    if (sum == b * b)
                    {
                        cnt++;
                    }
                }
            }
            Console.WriteLine(cnt);
        }
    }
}
// 文字列の入力
//string s = Console.ReadLine();

// 整数の入力
//long n = long.Parse(Console.ReadLine());

// 文字列配列の入力
//string[] inputStrArray = Console.ReadLine().Split(' ');

// 整数配列の入力
//long[] inputLongArray = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).ToArray();