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 = 400;
complex<long long> lis[MAXN];
int n;
int num[MAXN][MAXN];
int ans[MAXN];
long long cross(complex<long long> a, complex<long long> b) {
  return imag(conj(a) * b);
}
complex<long long> getPoint() {
  int x, y;
  scanf("%d%d", &x, &y);
  return complex<long long>(x, y);
}
int below(int i, int j) {
  return (real(lis[i]) == real(lis[j])) && (imag(lis[i]) < imag(lis[j]));
}
int betweenBelow(int i, int j, int x) {
  if (real(lis[i]) < real(lis[j])) {
    return real(lis[i]) < real(lis[x]) && real(lis[x]) < real(lis[j]) &&
           cross(lis[j] - lis[i], lis[x] - lis[i]) < 0;
  } else {
    return real(lis[j]) < real(lis[x]) && real(lis[x]) < real(lis[i]) &&
           cross(lis[i] - lis[j], lis[x] - lis[j]) < 0;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    lis[i] = getPoint();
  }
  memset(num, 0, sizeof(num));
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j)
      if (real(lis[i]) < real(lis[j])) {
        for (int k = (0); k < (n); ++k)
          if (k != i && k != j) {
            if (below(k, i)) num[i][j]++;
            if (below(k, j)) num[i][j]++;
            if (betweenBelow(i, j, k)) {
              num[i][j] += 2;
            }
          }
        num[j][i] = -num[i][j];
      }
  }
  memset(ans, 0, sizeof(ans));
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (i); ++j)
      for (int k = (0); k < (j); ++k) {
        int temp = abs(num[i][j] + num[j][k] + num[k][i]) / 2;
        temp -= betweenBelow(i, j, k);
        temp -= betweenBelow(j, k, i);
        temp -= betweenBelow(k, i, j);
        ans[temp]++;
      }
  long long x = (long long)n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  long long r = 0;
  for (int i = (0); i < (n - 2); ++i) r -= (long long)ans[i] * i * (n - 4);
  x += r / 2;
  for (int i = (0); i < (n - 2); ++i) x += (long long)i * (i - 1) / 2 * ans[i];
  printf("%lld\n", x);
  return 0;
}
