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)

#include <bits/stdc++.h>

#define EPS (1e-10)
#define rep(i, a, b) for (int i = a; i < (int)(b); ++i)
#define rrep(i, a, b) for (int i = b - 1; i >= (int)(a); --i)
#define all(a) a.begin(), a.end()

using namespace std;
using ll = long long;
using vi = vector<int>;
using vb = vector<bool>;
using vll = vector<ll>;
using vd = vector<double>;
using vvi = vector<vi>;
using vvb = vector<vb>;

constexpr int MOD = 1000000007;

bool check(const vi& a, const vi& b) {
    int d = 0;
    rep(i, 0, a.size()) { d += (a[i] - b[i]) * (a[i] - b[i]); }

    if (sqrt(d) - (int)sqrt(d) < EPS) { return true; }
    return false;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, d;
    cin >> n >> d;

    vvi p(n, vi(d, 0));
    rep(i, 0, n) {
        rep(j, 0, d) { cin >> p[i][j]; }
    }

    int cnt = 0;
    rep(i, 0, n) {
        rep(j, i + 1, n) {
            if (check(p[i], p[j])) { ++cnt; }
        }
    }
    cout << cnt << endl;

    return 0;
}