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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int d = s.nextInt();

		int i = 0;
		int j = 0;
		int k = 0;
		int sum = 0;//それぞれの距離の二乗
		int judgeSum = 0;//距離が整数なものの合計
		int[][] x = new int[n][d];
		for (i = 0; i < n; i++) {
			for (j = 0; j < d; j++) {//x
				x[i][j] = s.nextInt();
			}
			if (i != 0) {
				for (k = 0; k < i; k++) {//
					sum = 0;
					if (i != k) {
						//各要素を二乗して足す
						for (int m = 0; m < d; m++) {
							int sa = x[i][m] - x[k][m];
							int jijou = sa * sa;
							sum += jijou;
						}
					}
					judgeSum += (Math.sqrt(sum) * 10) % 10 == 0 ? 1 : 0;
				}

			}
		}

		System.out.println(judgeSum);

	}

}