You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.

Input

The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.

Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.

Output

Print the only integer c — the number of parallelograms with the vertices at the given points.

Example

Input

4
0 1
1 0
1 1
2 0


Output

1
#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 1e6 + 5;
const long long mod = 998244353;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long test = 1, n;
vector<pair<long long, long long> > v;
map<pair<long long, long long>, long long> ct;
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back({x, y});
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int mx = v[i].first + v[j].first, my = v[i].second + v[j].second;
      ct[{mx, my}]++;
    }
  }
  long long ans = 0;
  for (auto it : ct) {
    ans += (it.second - 1) * it.second / 2;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(15);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}
