This time our child has a simple polygon. He has to find the number of ways to split the polygon into non-degenerate triangles, each way must satisfy the following requirements:

  * each vertex of each triangle is one of the polygon vertex; 
  * each side of the polygon must be the side of exactly one triangle; 
  * the area of intersection of every two triangles equals to zero, and the sum of all areas of triangles equals to the area of the polygon; 
  * each triangle must be completely inside the polygon; 
  * each side of each triangle must contain exactly two vertices of the polygon. 



The picture below depicts an example of a correct splitting.

<image>

Please, help the child. Calculate the described number of ways modulo 1000000007 (109 + 7) for him.

Input

The first line contains one integer n (3 ≤ n ≤ 200) — the number of vertices of the polygon. Then follow n lines, each line containing two integers. The i-th line contains xi, yi (|xi|, |yi| ≤ 107) — the i-th vertex of the polygon in clockwise or counterclockwise order.

It's guaranteed that the polygon is simple.

Output

Output the number of ways modulo 1000000007 (109 + 7).

Examples

Input

4
0 0
0 1
1 1
1 0


Output

2


Input

4
0 0
1 0
0 1
-1 0


Output

1


Input

5
0 0
1 0
1 1
0 1
-2 -1


Output

3

Note

In the first sample, there are two possible splittings:

<image>

In the second sample, there are only one possible splitting:

<image>
#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, fl = 0;
long long f[205][205];
struct Poi {
  double x, y;
  double operator*(const Poi& A) const { return x * A.y - y * A.x; }
  Poi operator-(const Poi& A) const { return (Poi){x - A.x, y - A.y}; }
} p[205];
long long Solve(int l, int r) {
  if (f[l][r] != -1) return f[l][r];
  if (r - l == 1) return f[l][r] = 1;
  f[l][r] = 0;
  for (int i = l + 1; i <= r - 1; i++)
    if (((p[l] - p[r]) * (p[i] - p[r]) > 0))
      f[l][r] = (f[l][r] + Solve(l, i) * Solve(i, r) % 1000000007) % 1000000007;
  return f[l][r];
}
int main() {
  n = read();
  memset(f, -1, sizeof(f));
  for (int i = 1; i <= n; i++) p[i].x = read(), p[i].y = read();
  p[n + 1] = p[1];
  double now = 0;
  for (int i = 1; i <= n; i++) now += p[i] * p[i + 1];
  if (now < 0) reverse(p + 1, p + 1 + n);
  cout << Solve(1, n) << endl;
  return 0;
}
