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 = 303;
struct Point {
  long long x, y;
  pair<int, int> ind;
  Point(long long a = 0, long long b = 0, pair<int, int> c = {0, 0})
      : x(a), y(b), ind(c){};
  Point operator-(const Point& he) const {
    return Point(x - he.x, y - he.y, ind);
  }
  long long operator%(const Point& he) const { return x * he.y - y * he.x; }
  bool operator<(const Point& he) const {
    if (((*this).y > 0 or ((*this).y == 0 and (*this).x >= 0)) xor
        ((he).y > 0 or ((he).y == 0 and (he).x >= 0)))
      return ((*this).y > 0 or ((*this).y == 0 and (*this).x >= 0));
    return (*this) % he > 0;
  }
} in[maxN], T[maxN * maxN];
long long dp[maxN][maxN][6];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) scanf("%lld%lld", &in[i].x, &in[i].y);
  int m = 0;
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (n); j++)
      if (i != j) {
        T[m] = in[i] - in[j];
        T[m].ind = {i, j};
        m++;
      }
  sort(T, T + m);
  for (int i = (0); i < (n); i++) dp[i][i][0] = 1;
  for (int i = (0); i < (m); i++)
    for (int s = (0); s < (5); s++)
      for (int c = (0); c < (n); c++) {
        int a, b;
        tie(a, b) = T[i].ind;
        dp[c][a][s + 1] += dp[c][b][s];
      }
  long long res = 0;
  for (int i = (0); i < (n); i++) res += dp[i][i][5];
  printf("%lld\n", res);
  return 0;
}
