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 int MAXN = 333, K = 5, inf = 1023456789;
struct pt {
  int x, y, idx;
};
struct seg {
  pt a, b;
  double calc_atan2() const {
    int dx = b.x - a.x;
    int dy = b.y - a.y;
    return atan2((double)dy, (double)dx);
  }
  bool operator<(const seg& other) const {
    return calc_atan2() < other.calc_atan2();
  }
};
int N, x[MAXN], y[MAXN], dp[MAXN][MAXN][K + 5];
vector<seg> segs;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d %d", &x[i], &y[i]);
  }
  for (int i = 0; i < N; ++i) {
    for (int j = i + 1; j < N; ++j) {
      segs.push_back({{x[i], y[i], i}, {x[j], y[j], j}});
      segs.push_back({{x[j], y[j], j}, {x[i], y[i], i}});
    }
    dp[i][i][0] = 1;
  }
  sort(segs.begin(), segs.end());
  for (int i = 0; i < segs.size(); ++i) {
    int from = segs[i].a.idx;
    int to = segs[i].b.idx;
    for (int j = 0; j < N; ++j) {
      for (int k = 0; k < K; ++k) {
        dp[j][to][k + 1] += dp[j][from][k];
      }
    }
  }
  long long answer = 0;
  for (int i = 0; i < N; ++i) {
    answer += dp[i][i][K];
  }
  printf("%lld\n", answer);
  return 0;
}
