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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.Scanner

var N=0
var D=0
var answer=0
val sc=Scanner(System.`in`)
fun main(args:Array<String>){
    N=sc.nextInt()
    D=sc.nextInt()
    var ary=Array(N,{Array(D,{0})})
    for(i in 0..(N-1)){
        for(j in 0..(D-1)){
            ary[i][j]=sc.nextInt()
        }
    }
    for(i in 0..(N-1)){
        for(j in 0..(i-1)){
            var tmp=0
            for(k in 0..(D-1)){
                tmp+=(ary[i][k]-ary[j][k])*(ary[i][k]-ary[j][k])
            }
            if(tmp==root(tmp)*root(tmp)){
                answer++
            }
        }
    }
    println(answer)
}

fun root(a: Int): Int {
    var a = a

    var ketasu = 0
    var tmp = a
    while (tmp > 0) {
        ketasu++
        tmp /= 10
    }
    val constant = (ketasu + 1) / 2    //よく使うので定数化
    if (ketasu >= 1) {
        val suuji = IntArray(constant)
        tmp = a
        for (i in 0 until constant) {
            suuji[constant - 1 - i] = a % 100
            a /= 100
        }
        var ans = 0
        var kai = 0
        var mae = 0
        if (constant <= 1) {
            for (i in 10 downTo 0) {
                if (suuji[0] >= i * i) {
                    return i
                }
            }
        } else {
            for (i in 0 until constant) {
                mae += kai * 2
                for (j in 9 downTo 0) {
                    if ((mae * 10 + j) * j <= suuji[i]) {
                        ans = ans * 10 + j
                        mae *= 10
                        kai = j
                        if (i + 1 < constant) {
                            suuji[i + 1] += (suuji[i] - (mae + j) * j) * 100
                        }
                        break
                    }
                }
            }
        }
        return ans
    } else {
        return 0
    }
}