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;
struct Vector {
  long long x, y;
  Vector(long long a = 0, long long b = 0) : x(a), y(b) {}
};
Vector operator-(Vector a, Vector b) { return Vector(a.x - b.x, a.y - b.y); }
long long Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
bool operator<(Vector A, Vector B) { return Cross(A, B) < 0; }
const long long N = 305;
Vector p[N];
long long dp[N][N][6];
vector<pair<Vector, pair<long long, long long> > > e;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
signed main() {
  long long n = read();
  for (register long long i = (1); i <= (n); ++i)
    p[i].x = read(), p[i].y = read();
  for (register long long i = (1); i <= (n); ++i)
    for (register long long j = (1); j <= (n); ++j)
      if (i ^ j) e.push_back(make_pair(Vector(p[j] - p[i]), make_pair(i, j)));
  sort(e.begin(), e.end());
  for (auto now : e) {
    long long u = now.second.first, v = now.second.second;
    dp[u][v][1]++;
    for (register long long i = (1); i <= (5); ++i)
      for (register long long s = (1); s <= (n); ++s)
        dp[s][v][i + 1] += dp[s][u][i];
  }
  long long ans = 0;
  for (register long long i = (1); i <= (n); ++i) ans += dp[i][i][5];
  printf("%lld\n", ans);
  return 0;
}
