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 mod = 1e9 + 7;
struct fastio {
  char s[100000];
  int it, len;
  fastio() { it = len = 0; }
  inline char get() {
    if (it < len) return s[it++];
    it = 0;
    len = fread(s, 1, 100000, stdin);
    if (len == 0)
      return EOF;
    else
      return s[it++];
  }
  bool notend() {
    char c = get();
    while (c == ' ' || c == '\n') c = get();
    if (it > 0) it--;
    return c != EOF;
  }
} _buff;
inline long long getnum() {
  long long r = 0;
  bool ng = 0;
  char c;
  c = _buff.get();
  while (c != '-' && (c < '0' || c > '9')) c = _buff.get();
  if (c == '-') ng = 1, c = _buff.get();
  while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = _buff.get();
  return ng ? -r : r;
}
template <class T>
inline void putnum(T x) {
  if (x < 0) putchar('-'), x = -x;
  register short a[20] = {}, sz = 0;
  while (x) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline char getreal() {
  char c = _buff.get();
  while (c <= 32) c = _buff.get();
  return c;
}
long long qpow(long long x, long long k) {
  return k == 0 ? 1
                : 1ll * qpow(1ll * x * x % mod, k >> 1) * (k & 1 ? x : 1) % mod;
}
struct point {
  long long x, y;
  point(long long X = 0, long long Y = 0) {
    x = X;
    y = Y;
  }
  point operator+(const point &t) const { return point(x + t.x, y + t.y); }
  point operator-(const point &t) const { return point(x - t.x, y - t.y); }
  point operator*(const point &t) const {
    return point(x * t.x - y * t.y, x * t.y + y * t.x);
  }
  long long len() const { return x * x + y * y; }
  long long det(const point &t) const { return x * t.y - y * t.x; }
  long long dot(const point &t) const { return x * t.x + y * t.y; }
  bool operator<(const point &t) const {
    return make_pair(x, y) < make_pair(t.x, t.y);
  }
  long long detll(const point &t) const {
    double tmp = 1.0 * x * t.y - 1.0 * y * t.x;
    if (tmp > 1e18) return 1e18;
    if (tmp < -1e18) return -1e18;
    return x * t.y - y * t.x;
  }
  long long dotll(const point &t) const {
    double tmp = 1.0 * x * t.y + 1.0 * y * t.x;
    if (tmp > 1e18) return 1e18;
    if (tmp < -1e18) return -1e18;
    return x * t.y + y * t.x;
  }
};
int n;
point a[311];
long long all;
point O;
mt19937 rnd(time(0));
void getO() {
  while (true) {
    O = point(rnd() % 1000000000, rnd() % 1000000000);
    bool ok = true;
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++) {
        if ((a[i] - O).det(a[j] - O) == 0) ok = false;
      }
    if (ok) break;
  }
}
int get(int i, int j, int k) {
  int ret = 0;
  ret += (a[i] - O).det(a[k] - O) > 0 && (a[k] - O).det(a[j] - O) > 0 &&
         (a[k] - a[i]).det(a[j] - a[i]) > 0;
  ret -= (a[i] - O).det(a[k] - O) < 0 && (a[k] - O).det(a[j] - O) < 0 &&
         (a[k] - a[i]).det(a[j] - a[i]) < 0;
  return ret;
}
long long getC3() {
  static long long tab[311][311];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        tab[i][j] += get(i, j, k);
      }
    }
  }
  long long ret = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (i == j || i == k || j == k) continue;
        int cnt = tab[i][j] + tab[j][k] + tab[k][i];
        cnt -= get(i, j, k) + get(j, k, i) + get(k, i, j);
        cnt = abs(cnt);
        ret += 1ll * cnt * (cnt - 1) / 2;
      }
    }
  }
  return ret / 6;
}
long long getC4() {
  long long ret = 1ll * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 24;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      int cnt = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        cnt += (a[j] - a[i]).det(a[k] - a[i]) > 0;
      }
      ret -= 1ll * cnt * (cnt - 1) * (cnt - 2) / 6;
    }
  }
  return ret;
}
int main() {
  n = getnum();
  for (int i = 1; i <= n; i++) {
    a[i].x = getnum(), a[i].y = getnum();
  }
  all = 1ll * n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 120;
  long long ans = all;
  getO();
  long long C3 = getC3();
  long long C4 = getC4();
  cout << ans - C4 + C3 << endl;
  return 0;
}
