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++14 (GCC 5.4.1) (judge verdict: Accepted)

/****************************************************
*   Template for coding contests                    *
*   Author    :    Sanjeev Sharma                   *
*   Email     :    thedevelopersanjeev@gmail.com    *
*****************************************************/
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;

bool f(vector<double> &pointA, vector<double> pointB){
    double dist = 0.0;
    int n = pointA.size();
    double temp = 0.0;
    for(int i = 0; i < n; i++)
        temp += (abs(pointA[i]-pointB[i]) * abs(pointA[i]-pointB[i]));
    dist = sqrt(temp);
    return ceil(dist) == floor(dist);
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, d;
    cin >>n >>d;
    vector<vector<double>> points(n);
    for(int i = 0; i < n; i++){
        vector<double> point(d);
        for(int j = 0; j < d; j++)
            cin >>point[j];
        points[i] = point;
    }
    int ans = 0;
    for(int i = 0; i < n-1; i++){
        for(int j = i+1; j < n; j++){
            if(f(points[i], points[j]))
                ans++;
        }
    }
    cout <<ans;
    return 0;
}