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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;

use std::str::FromStr;

 

fn read<T: FromStr>() -> T {

    let stdin = stdin();

    let stdin = stdin.lock();

    let token: String = stdin

        .bytes()

        .map(|c| c.expect("failed to read char") as char)

        .skip_while(|c| c.is_whitespace())

        .take_while(|c| !c.is_whitespace())

        .collect();

    token.parse().ok().expect("failed to parse token")

}

fn main() {

    //点

    let n: usize = read();

    //次元

    let d: usize = read();

    let mut x: [[i32; 10]; 10] = [[0; 10]; 10];

    let mut ans: u32 = 0;

 

    for i in 0..n {

        for j in 0..d {

            x[i][j] = read();

        }

    }

 

    for i in 0..n {

        for j in i + 1..n {

            let mut sum = 0;

            for k in 0..d {

                sum += (x[i][k] - x[j][k]).abs().pow(2);

            }

            let distance = (sum as f64).sqrt();

            if distance % 1.0 == 0.0 {

                ans += 1;

            }

        }

    }

 

    println!("{:?}", ans);

}