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;
const double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void ckmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T& a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const int maxn = 300 + 10;
int n;
int x[maxn], y[maxn];
int dir[maxn][maxn][maxn];
long long f1[maxn][maxn][4];
long long f2[maxn][maxn][4];
int outer(int a, int b, int c) {
  long long t = (long long)(x[b] - x[a]) * (y[c] - y[a]) -
                (long long)(x[c] - x[a]) * (y[b] - y[a]);
  if (t == 0) return 0;
  return (t < 0) ? -1 : 1;
}
int main() {
  std::ios::sync_with_stdio(false);
  while (cin >> n) {
    for (int i = 0; i < (n); ++i) cin >> x[i] >> y[i];
    for (int i = 0; i < (n); ++i)
      for (int j = (i + 1); j < (n); ++j)
        if (x[j] < x[i] || x[j] == x[i] && y[j] < y[i]) {
          swap(x[i], x[j]);
          swap(y[i], y[j]);
        }
    for (int i = 0; i < (n); ++i)
      for (int j = 0; j < (n); ++j)
        for (int k = 0; k < (n); ++k) dir[i][j][k] = outer(i, j, k);
    long long ret = 0;
    vector<vector<int>> prev(n);
    vector<vector<int>> next(n);
    for (int i = 0; i < (n); ++i) {
      for (int j = 0; j < (i); ++j) prev[i].push_back(j);
      for (int j = (i + 1); j < (n); ++j) next[i].push_back(j);
      sort(prev[i].begin(), prev[i].end(),
           [&](int a, int b) { return dir[i][a][b] > 0; });
      sort(next[i].begin(), next[i].end(),
           [&](int a, int b) { return dir[i][a][b] > 0; });
    }
    for (int src = 0; src < (n); ++src) {
      memset(f1, 0, sizeof(f1));
      memset(f2, 0, sizeof(f2));
      for (int i = (src + 1); i < (n); ++i) {
        f1[src][i][0] = 1;
        f2[src][i][0] = 1;
        auto& a = prev[i];
        auto& b = next[i];
        long long s1[4], s2[4];
        memset(s1, 0, sizeof(s1));
        memset(s2, 0, sizeof(s2));
        for (int pa = 0; pa < (((int)a.size())); ++pa)
          for (int c = 0; c < (3); ++c) s2[c] += f2[a[pa]][i][c];
        for (int pb = 0, pa = 0; pb < ((int)b.size()); pb++) {
          for (; pa < ((int)a.size()) && dir[a[pa]][i][b[pb]] > 0; ++pa) {
            for (int c = 0; c < (3); ++c) s1[c] += f1[a[pa]][i][c];
            for (int c = 0; c < (3); ++c) s2[c] -= f2[a[pa]][i][c];
          }
          for (int c = 0; c < (3); ++c) f1[i][b[pb]][c + 1] += s1[c];
          for (int c = 0; c < (3); ++c) f2[i][b[pb]][c + 1] += s2[c];
        }
        memset(s1, 0, sizeof(s1));
        memset(s2, 0, sizeof(s2));
        for (int j = (src); j < (i); ++j)
          for (int c = 0; c < (4); ++c) {
            s1[c] += f1[j][i][c];
            s2[c] += f2[j][i][c];
          }
        for (int c1 = 0; c1 < (4); ++c1)
          for (int c2 = 0; c2 < (4); ++c2)
            if (c1 + c2 == 3) ret += s1[c1] * s2[c2];
      }
      for (int i = (src + 1); i < (n); ++i) {
        int k = 0;
        for (; k < ((int)prev[i].size()) && prev[i][k] != src; ++k)
          ;
        if (k >= ((int)prev[i].size())) continue;
        for (int j = (k); j < (((int)prev[i].size()) - 1); ++j)
          prev[i][j] = prev[i][j + 1];
        prev[i].pop_back();
      }
    }
    cout << ret << endl;
    break;
  }
  return 0;
}
