10
-2 1
2 -2
-1 -2
0 0
2 -1
0 -2
2 2
0 2
-1 -1
1 -2
 40
-8 24
2 -1
1 -18
72 -70
5 -4
-308 436
-19 40
36 -35
-178 265
-1 2
-7 30
-1 0
3 -2
200 -285
17 -16
-35 74
0 -4
-86 106
-1 4
-7 6
0 1
-5 4
-2 3
6 -5
-4 5
181 -262
76 -118
0 0
-7 18
-58 104
-5 6
-6 12
-3 4
1 0
11 -10
-86 130
-3 6
153 -236
-183 270
-33 64
 5
0 -1
0 -2
0 -3
0 -4
0 -5
 10
0 1
-1 -1
1 1
-1 0
1 -1
-2 -1
-2 2
-2 0
0 -2
0 -1
 20
-1 18
-2 5
-5 4
2 -33
9 -18
0 0
11 -22
2 0
-1 2
-4 41
1 6
1 -2
6 -12
0 1
-3 6
3 -6
3 -8
-1 4
2 -5
1 0
 10
2 1
-1 0
-2 -1
-1 1
0 2
2 -2
0 0
-2 -2
0 -2
-2 1
 20
-3 -3
0 4
-3 1
1 1
-1 2
-4 4
3 -1
-3 0
0 2
4 0
2 3
2 4
4 -3
-4 3
-1 1
1 3
-2 4
1 -2
1 -1
3 0
 3
0 0
0 -1
0 1
 10
2 1
-1 1
0 0
-3 1
-2 -3
-1 -2
-1 -1
1 2
3 -2
0 -2
 3
3 3
1 2
1 1
 4
0 0
1 0
2 0
3 0
 9
1 1
1 0
0 1
0 0
-1 0
-1 1
-1 -1
1 -1
0 -1
 7
1 1
-1 -1
1 0
0 1
-1 0
0 -1
0 0

4
 57
 4
 5
 21
 4
 10
 1
 1
 0
 2
 8
 3

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int N = 2e3 + 7, M = 1e4, OO = 0x3f3f3f3f;
int vis[N][N];
int main() {
  int TCF = 1;
  while (TCF--) {
    int n;
    scanf("%d", &n);
    ;
    vector<pair<int, int>> p;
    int x, y;
    for (int i = 0; i < n; ++i) {
      scanf("%d %d", &x, &y);
      p.emplace_back(x, y);
      vis[x + 1000][y + 1000] = 1;
    }
    vector<pair<int, int>> mid;
    int ans = 0;
    for (int i = (0); i < (n); i++) {
      for (int j = (i + 1); j < (n); j++) {
        if (i == j) continue;
        int mx = p[i].first + p[j].first;
        int my = p[i].second + p[j].second;
        if (mx % 2 == 0 && my % 2 == 0) {
          mx /= 2;
          my /= 2;
          if (vis[mx + 1000][my + 1000]) {
            ans++;
          }
        }
      }
    }
    printf("%d\n", ans);
  }
}
