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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
/**
 * ABC133
 * B - Good Distance
 */

// $s = trim(fgets(STDIN));
// fscanf(STDIN, "%d %d", $a, $b);
// $n = array_map('intval', explode(' ', trim(fgets(STDIN))));

fscanf(STDIN, "%d %d", $N, $D);
for ($i = 0; $i < $N; $i++) {
    $X[$i] = array_map('intval', explode(' ', trim(fgets(STDIN))));
}

$count = 0;
for ($i = 0; $i < $N - 1; $i++) {
    for ($j = $i + 1; $j < $N; $j++) {
        $distance = 0;
        for ($k = 0; $k < $D; $k++) {
            $distance += pow(abs($X[$i][$k] - $X[$j][$k]), 2);
        }
        if (is_pythagoras($distance)) $count++;
    }
}

echo $count, PHP_EOL;

function is_pythagoras($number) {
    for ($i = 0; pow($i, 2) <= $number; $i++) {
        if (pow($i, 2) === $number) return true;
    }
    return false;
}