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 static System.Console;
using static System.Math;
using System.IO;
using System.Collections.Generic;
namespace I_am_the_green_coder{
    class Inonoa{
        // なんか入力用の関数作ったけどまだテストしてないので使う前に確かめて
        static void Main(){
            long[] nd = ILongs();

            long[][] x = ILongLongs(nd[0]);

            long ans = 0;

            for(long i=0;i<nd[0]-1;i++){
                for(long j=i+1;j<nd[0];j++){
                    long d2 = 0;
                    for(int d=0;d<nd[1];d++){
                        d2 += (x[i][d]-x[j][d])*(x[i][d]-x[j][d]);
                    }

                    for(long r=0;r<=(long)Math.Sqrt(d2);r++){
                        if(r*r==d2){
                            ans ++;
                            break;
                        }
                    }
                    
                }
            }

            Console.WriteLine(ans);
        }

        static readonly long MOD = 1000000007;

        ///<summary>a_1, a_2, a_3, ...みたいなの</summary>
        static long[] ILongs(){
            string[] s = ReadLine().Split();
            long[] re = new long[s.Length];
            for(long i=0;i<s.Length;i++){
                re[i] = long.Parse(s[i]);
            }
            return re;
        }

        ///<summary>xooxoxoxxoxo...みたいなの</summary>
        static char[] IChars(){
            return ReadLine().ToCharArray();
        }

        ///<summary>n m とかをちょっとはやくかくだけ</summary>
        static string[] IStrings(){
            return ReadLine().Split();
        }

        ///<summary>n行の入力を取ったり</summary>
        static long[][] ILongLongs(long n){
            long[][] re = new long[n][];
            for(long i=0;i<n;i++){
                string[] s = ReadLine().Split();
                re[i] = new long[s.Length];
                for(long j=0;j<s.Length;j++){
                    re[i][j] = long.Parse(s[j]);
                }
            }
            return re;
        }

        ///<summary>n行の文字列を取ったり</summary>
        static char[][] ICharsChars(long n){
            char[][] re = new char[n][];
            for(long i=0;i<n;i++){
                re[i] = ReadLine().ToCharArray();
            }
            return re;
        }
    }
}