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;
int first[300], second[300];
pair<int, int> v[90000];
long long int cross(pair<int, int> a, pair<int, int> b) {
  return (long long int)a.first * b.second - (long long int)a.second * b.first;
}
int comp(pair<int, int> a, pair<int, int> b) {
  pair<int, int> va = make_pair(first[a.second] - first[a.first],
                                second[a.second] - second[a.first]);
  pair<int, int> vb = make_pair(first[b.second] - first[b.first],
                                second[b.second] - second[b.first]);
  int ha = (va.second > 0) || ((va.second == 0) && (va.first > 0));
  int hb = (vb.second > 0) || ((vb.second == 0) && (vb.first > 0));
  if (ha && !hb)
    return 1;
  else if (!ha & hb)
    return 0;
  else if (cross(va, vb) > 0)
    return 1;
  else
    return 0;
}
long long int dp[300][300][6];
int main() {
  int i;
  int n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d %d", &first[i], &second[i]);
  int j, k, c = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i != j) v[c++] = make_pair(i, j);
    }
  }
  sort(v, v + c, comp);
  for (i = 0; i < n; i++) dp[i][i][0] = 1;
  for (i = 0; i < c; i++) {
    for (j = 0; j < n; j++) {
      for (k = 0; k < 5; k++) dp[j][v[i].second][k + 1] += dp[j][v[i].first][k];
    }
  }
  long long int ans = 0;
  for (i = 0; i < n; i++) ans += dp[i][i][5];
  printf("%I64d\n", ans);
  return 0;
}
