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;
using ll = long long;
using ld = long double;
const ll MOD = 998244353;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
ll modInv(ll a) { return binpow(a, MOD - 2); }
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 301;
pair<int, int> arr[mxN];
int n;
int half(pair<int, int> a) {
  return a.second == 0 ? a.first < 0 : a.second > 0;
}
ll cross(pair<int, int> a, pair<int, int> b) {
  return 1LL * a.first * b.second - 1LL * b.first * a.second;
}
bool comp(pair<int, int> a, pair<int, int> b) {
  int A = half(a);
  int B = half(b);
  return A == B ? cross(a, b) > 0 : A < B;
}
ll nCr(int n, int r) {
  ll res = 1;
  for (int i = (0); i < (r); ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
bool ok(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  return (cross(a, b) + cross(b, c) + cross(c, a)) > 0;
}
bitset<mxN> inside[mxN][mxN];
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    cin >> arr[i].first >> arr[i].second;
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      if (i == j) continue;
      for (int k = (0); k < (n); ++k) {
        if (k == j || k == i) continue;
        inside[i][j][k] = ok(arr[i], arr[j], arr[k]);
      }
    }
  }
  ll ans = 0;
  for (int i = (0); i < (n); ++i) {
    for (int j = (i + 1); j < (n); ++j) {
      for (int k = (i + 1); k < (n); ++k) {
        int num = (inside[i][j] & inside[j][k] & inside[k][i]).count();
        ans -= (nCr(num, 2));
      }
    }
  }
  for (int i = (0); i < (n); ++i) {
    pair<int, int> curr = arr[i];
    int save = 0;
    vector<pair<int, int> > nxt;
    for (int j = (0); j < (n); ++j) {
      if (i != j) {
        nxt.push_back({arr[j].first - curr.first, arr[j].second - curr.second});
      }
    }
    sort(nxt.begin(), nxt.end(), comp);
    int last = 0;
    for (int j = (0); j < (n - 1); ++j) {
      while (last < j + n - 1 && cross(nxt[j], nxt[last % (n - 1)]) >= 0)
        last++;
      int right = last - j - 1;
      ans += max(nCr(right, 2) * (n - right - 2), (ll)0);
      save += max(nCr(right, 2) * (n - right - 2), (ll)0);
    }
  }
  ll all = nCr(n, 5) * 6 - ans;
  cout << all << "\n";
  return 0;
}
