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 INF = 1e9 + 10;
const int SZ = 5e5 + 10;
const int mod = 1e9 + 7;
const double PI = acos(-1);
const double eps = 1e-7;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
struct haha {
  long long x, y;
  int id;
} a[SZ], e;
haha operator-(haha a, haha b) { return (haha){a.x - b.x, a.y - b.y}; }
long long operator*(haha a, haha b) { return a.x * b.y - a.y * b.x; }
bool cmp(haha a, haha b) { return (a - e) * (b - e) > 0; }
bitset<310> b[310][310], tmp;
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    a[i].x = read(), a[i].y = read();
    a[i].id = i;
    tmp.set(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        if ((a[k] - a[i]) * (a[j] - a[i]) < 0) b[i][j].set(k);
      }
    }
  }
  long long ans = 1ll * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120,
            sum4 = 0, sum3 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (i == j || j == k || i == k) continue;
        bitset<310> now1 = tmp, now4 = tmp;
        now1 &= b[i][j];
        now1 &= b[k][j];
        now1 &= b[i][k];
        now4 &= b[j][i];
        now4 &= b[i][k];
        now4 &= b[k][j];
        long long n1 = now1.count(), n4 = now4.count();
        long long d3 = 0, d4 = 0;
        d3 += n4 * (n4 - 1) / 2;
        d4 += n1 * n4;
        sum3 += d3;
        sum4 += d4;
      }
    }
  }
  sum3 /= 3;
  sum4 /= 2;
  ans -= sum3 + sum4;
  cout << ans << endl;
}
