You are given a set of n points in a 2D plane. No three points are collinear.

A pentagram is a set of 5 points A,B,C,D,E that can be arranged as follows. <image> Note the length of the line segments don't matter, only that those particular intersections exist.

Count the number of ways to choose 5 points from the given set that form a pentagram.

Input

The first line contains an integer n (5 ≤ n ≤ 300) — the number of points.

Each of the next n lines contains two integers x_i, y_i (-10^6 ≤ x_i,y_i ≤ 10^6) — the coordinates of the i-th point. It is guaranteed that no three points are collinear.

Output

Print a single integer, the number of sets of 5 points that form a pentagram.

Examples

Input


5
0 0
0 2
2 0
2 2
1 3


Output


1


Input


5
0 0
4 0
0 4
4 4
2 3


Output


0


Input


10
841746 527518
595261 331297
-946901 129987
670374 -140388
-684770 309555
-302589 415564
-387435 613331
-624940 -95922
945847 -199224
24636 -565799


Output


85

Note

A picture of the first sample: <image> A picture of the second sample: <image> A picture of the third sample: <image>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pt = pair<ll, ll>;
inline pt operator-(const pt& a, const pt& b) {
  return pt{a.first - b.first, a.second - b.second};
}
inline ll cross(pt a, pt b) { return a.first * b.second - a.second * b.first; }
inline ll cross(pt a, pt b, pt c) { return cross(b - a, c - a); }
inline bool dir(pt a, pt b, pt c) {
  ll v = cross(a, b, c);
  assert(v != 0);
  return v > 0;
}
const int MAXN = 310;
using bs = bitset<300>;
int N;
pt P[MAXN];
bs D[MAXN][MAXN];
ll numDir[MAXN][MAXN][2];
ll numCross[MAXN][MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> P[i].first >> P[i].second;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      for (int k = 0; k < N; k++) {
        if (k == i || k == j) continue;
        bool d = dir(P[i], P[j], P[k]);
        D[i][j][k] = d;
        numDir[i][j][d]++;
      }
    }
  }
  using ll = long long;
  ll res = ll(N) * ll(N - 1) * ll(N - 2) * ll(N - 3) * ll(N - 4) / 2 / 6;
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      for (int k = j + 1; k < N; k++) {
        bs ij = D[i][j];
        bs jk = D[j][k];
        bs ki = D[k][i];
        if (!D[i][j][k]) {
          ij = D[j][i];
          jk = D[k][j];
          ki = D[i][k];
        }
        ij[i] = ij[j] = ij[k] = false;
        jk[i] = jk[j] = jk[k] = false;
        ki[i] = ki[j] = ki[k] = false;
        int c = (ij & jk & ki).count();
        numCross[i][k][j] = (ij & jk).count() - c;
        numCross[j][k][i] = (ij & ki).count() - c;
        numCross[i][j][k] = (jk & ki).count() - c;
        res -= ll(c) * ll(N - 3 - c);
      }
    }
  }
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      for (int k = 0; k < N; k++) {
        if (k == i || k == j) continue;
        res -= ll(numCross[i][j][k]) * ll(numCross[i][j][k] - 1) / 2;
      }
    }
  }
  ll res2 = 1ll * N * (N - 1) * (N - 2) * (N - 3) * (N - 4) / 120;
  ll res3 = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      ll a = numDir[i][j][0];
      res3 += a * (a - 1) * (a - 2) / 6;
    }
  }
  ll ans = res - 14 * res2 + 2 * res3;
  cout << ans << '\n';
  return 0;
}
