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 (GCC 5.4.1) (judge verdict: Accepted)

// AtCoder ABC133: B - Good Distance
// 2019.9.2 bal4u

#include <stdio.h>
#include <math.h>

#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif

int in()   // 整数の入力（負数対応）
{
	int n = 0, c = gc();
	if (c == '-') {	c = gc();
		do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
		return -n;
	}
	do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

int x[25][25];

int main()
{
	int N, D, r, rr, c, s, b, ans;
	
	N = in(), D = in();
	for (r = 0; r < N; r++) for (c = 0; c < D; c++) x[r][c] = in();
	ans = 0;
	for (r = 0; r < N; r++) for (rr = r+1; rr < N; rr++) {
		s = 0; for (c = 0; c < D; c++) s += (x[r][c]-x[rr][c])*(x[r][c]-x[rr][c]);
		b = (int)sqrt((double)s);
		if (b*b == s) ans++;
	}
	printf("%d\n", ans);
	return 0;
}