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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ull = unsigned long long;
using ld = long double;

#define MOD 1000000007LL

#define rep(i, n) for(ll (i) = 0LL;(i) < (ll)(n);(i)++)
#define rep2(i, s, e) for(ll (i) = (ll)(s);(i) < (ll)(e);(i)++)
#define repi(i, n) for(ll (i) = 0LL;(i) <= (ll)(n);(i)++)
#define repi2(i, s, e) for(ll (i) = (ll)(s);(i) <= (ll)(e);(i)++)
#define per(i, n) for(ll (i) = (ll)(n) - 1LL;(i) >= 0LL;(i)--)
#define per2(i, s, e) for(ll (i) = (ll)(s) - 1LL;(i) >= (ll)(e);(i)--)
#define peri(i, n) for(ll (i) = (ll)(n);(i) >= 0LL;(i)--)
#define peri2(i, s, e) for(ll (i) = (ll)(s);(i) >= (ll)(e);(i)--)
#define iter(i, it) for(auto &(i): (it))

template<typename T, typename U> ostream& operator<<(ostream &s, const pair<T, U> m) {
    cout << "(" << m.first << ", " << m.second << ")";
    return s;
}
template<typename T, typename U> ostream& operator<<(ostream &s, const map<T, U> m) {
    ll c = 0;
    cout << "{ ";
    iter(i, m) cout << i << (c++ == m.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}
template<typename T> ostream& operator<<(ostream &s, const vector<T> &v) {
    cout << "{ ";
    rep(i, v.size()) cout << v[i] << (i == v.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}
template<typename T> ostream& operator<<(ostream &s, const list<T> &v) {
    ll c = 0;
    cout << "{ ";
    iter(i, v) cout << i << (c++ == v.size() - 1 ? " " : ", ");
    cout << "}";
    return s;
}

int main(void) {
    ll N, D;
    cin >> N >> D;
    vector<vector<ll>> X(N, vector<ll>(D));
    ll ans = 0;
    rep(i, N) {
        rep(j, D) {
            cin >> X[i][j];
        }
    }
    rep(i, N - 1) {
        rep2(j, i + 1, N) {
            ll temp = 0;
            rep(x, D) {
                ll tmp = X[i][x] - X[j][x];
                temp += tmp * tmp;
            }
            ld tmp2 = sqrtl(temp);
            if(tmp2 == floorl(tmp2)) ans++;
        }
    }
    cout << ans << endl;
    return 0;
}