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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double PI = acos(-1);
const long double eps = 1e-6;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 310, LOG = 20;
long long n, m, k, t, a, b, ans;
long double ang[MAXN][MAXN];
long double X[MAXN], Y[MAXN];
long long dp[MAXN][MAXN][6];
pair<int, int> A[MAXN * MAXN];
int f(long double x, long double y) {
  if (abs(y) < eps) {
    if (x > 0) return 0;
    return 2;
  }
  if (y > 0) return 1;
  return 2;
}
bool cmp(pair<int, int> i, pair<int, int> j) {
  long double a1 = ang[i.first][i.second], a2 = ang[j.first][j.second];
  long double x1 = X[i.second] - X[i.first], y1 = Y[i.second] - Y[i.first];
  long double x2 = X[j.second] - X[j.first], y2 = Y[j.second] - Y[j.first];
  int f1 = f(x1, y1), f2 = f(x2, y2);
  if (f1 != f2) return f1 < f2;
  return y1 * x2 < y2 * x1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> X[i] >> Y[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) {
        long double x = X[j] - X[i], y = Y[j] - Y[i];
        long double z = sqrt(x * x + y * y);
        long double sn = y / z;
        long double theta = asin(sn) * 180 / PI;
        ang[i][j] = theta;
        A[++t] = {i, j};
      }
  sort(A + 1, A + t + 1, cmp);
  for (int i = 0; i < n; i++) dp[i][i][0] = 1;
  for (int e = 1; e <= t; e++) {
    int u = A[e].first, v = A[e].second;
    for (int cnt = 1; cnt <= 5; cnt++)
      for (int i = 0; i < n; i++) dp[i][v][cnt] += dp[i][u][cnt - 1];
  }
  for (int i = 0; i < n; i++) ans += dp[i][i][5];
  cout << ans << '\n';
  return 0;
}
