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>
struct op {
  int u, v;
  long double c;
};
using namespace std;
const int M = 200005, N = 305;
const long double pi = 3.1415926535;
int i, j, n, k, a[N][2], so, bc, l, r, mid, ot;
long long f[N][N][6], ans;
int bh[N][N];
op lk[M];
long double ch;
int read() {
  int ef = 1, x = 0;
  char s = getchar();
  while ((s < '0') || (s > '9')) {
    if (s == '-') ef = -1;
    s = getchar();
  }
  while ((s >= '0') && (s <= '9')) x = x * 10 + (s - '0'), s = getchar();
  return ef * x;
}
void swap(int x, int y) {
  lk[0].u = lk[x].u;
  lk[x].u = lk[y].u;
  lk[y].u = lk[0].u;
  lk[0].v = lk[x].v;
  lk[x].v = lk[y].v;
  lk[y].v = lk[0].v;
  lk[0].c = lk[x].c;
  lk[x].c = lk[y].c;
  lk[y].c = lk[0].c;
}
void kp(int l, int r) {
  int x = l, y = r;
  long double mid = lk[(x + y) / 2].c;
  while (x <= y) {
    while (mid > lk[x].c) x++;
    while (lk[y].c > mid) y--;
    if (x <= y) {
      swap(x, y);
      x++;
      y--;
    }
  }
  if (l < y) kp(l, y);
  if (x < r) kp(x, r);
}
long double js(int x, int y) {
  return atan2(a[y][1] - a[x][1], a[y][0] - a[x][0]) / pi * 180;
}
int main() {
  n = read();
  for (i = 1; i <= n; i++) {
    a[i][0] = read(), a[i][1] = read();
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (i != j) {
        so++;
        lk[so].u = i;
        lk[so].v = j;
        lk[so].c = js(i, j);
        atan2(a[j][1] - a[i][1], a[j][0] - a[i][0]);
        bh[i][j] = so;
      }
    }
  }
  kp(1, so);
  for (i = 1; i <= so; i++) {
    f[lk[i].u][lk[i].v][1] = 1;
    for (k = 1; k <= 4; k++) {
      for (j = 1; j <= n; j++) {
        f[j][lk[i].v][k + 1] += f[j][lk[i].u][k];
      }
    }
  }
  for (i = 1; i <= n; i++) {
    ans += f[i][i][5];
  }
  printf("%lld\n", ans);
  fclose(stdin);
  fclose(stdout);
}
