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;
struct Pt {
  long long x, y;
  Pt() {}
  Pt(long long _x, long long _y) : x(_x), y(_y) {}
  Pt operator+(const Pt& p) const { return Pt(x + p.x, y + p.y); }
  Pt operator-(const Pt& p) const { return Pt(x - p.x, y - p.y); }
  Pt operator*(long long c) const { return Pt(x * c, y * c); }
  Pt operator/(long long c) const { return Pt(x / c, y / c); }
  bool operator<(const Pt& p) const { return x == p.x ? y < p.y : x < p.x; }
  long long abs2() { return x * x + y * y; }
};
long long dp[305][305][6];
int main() {
  int n;
  scanf("%d", &n);
  vector<Pt> p(n);
  struct Seg {
    Pt p;
    int s, t;
  };
  vector<Seg> segs;
  for (int i = 0; i < n; ++i) {
    scanf("%lld%lld", &p[i].x, &p[i].y);
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i != j) {
        Pt seg = p[j] - p[i];
        segs.push_back({seg, i, j});
      }
  sort(segs.begin(), segs.end(), [&](Seg aa, Seg bb) {
    Pt a = aa.p, b = bb.p;
    assert((a.x || a.y) && (b.x || b.y));
    if (a.y <= 0 && b.y > 0) return true;
    if (a.y > 0 && b.y <= 0) return false;
    if (!a.y && !b.y) return a.x < b.x;
    return a.x * b.y > b.x * a.y;
  });
  for (int i = 0; i < n; ++i) dp[i][i][0] = 1;
  for (auto& seg : segs) {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < 5; ++j) {
        dp[i][seg.t][j + 1] += dp[i][seg.s][j];
      }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += dp[i][i][5];
  printf("%lld\n", ans);
}
