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 {
  pair<int, int> id;
  long long int x, y;
  pt() {}
  pt(pt p, int id1, int id2) {
    x = p.x;
    y = p.y;
    id.first = id1;
    id.second = id2;
  }
  pt(long long int x_, long long int y_) { x = x_, y = y_; }
  pt operator+(const pt &p) { return {x + p.x, y + p.y}; }
  pt operator-(const pt &p) { return {x - p.x, y - p.y}; }
  pt operator*(const long long int &t) { return {x * t, y * t}; }
  long long int operator*(const pt &p) { return x * p.x + y * p.y; }
  long long int operator^(const pt &p) { return x * p.y - y * p.x; }
};
bool half(pt p) {
  assert(p.x != 0 || p.y != 0);
  return p.y > 0 || (p.y == 0 && p.x < 0);
}
void polarSort(vector<pt> &v) {
  sort(v.begin(), v.end(), [](pt v, pt w) {
    return make_pair(half(v), 0LL) < make_pair(half(w), (v ^ w));
  });
}
vector<pt> edges;
vector<pt> vp;
long long int dp[305][305][6];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long int x, y;
    scanf("%lld %lld", &x, &y);
    vp.push_back(pt(x, y));
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      edges.push_back(pt(vp[i] - vp[j], j, i));
    }
  }
  polarSort(edges);
  for (int i = 0; i < n; i++) dp[i][i][0] = 1;
  for (int i = 0; i < (int)edges.size(); i++) {
    int st = edges[i].id.first, en = edges[i].id.second;
    for (int l = 5; l >= 1; l--) {
      for (int j = 0; j < n; j++) {
        dp[j][en][l] += dp[j][st][l - 1];
      }
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) ans += dp[i][i][5];
  printf("%lld\n", ans);
  return 0;
}
