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>
#pragma comment(linker, "/STACK:100000000000,100000000000")
using namespace std;
const long long inf = 2e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-7;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
int n, x[205], y[205];
long long mul(int a, int b, int c) {
  return 1LL * (x[b] - x[a]) * (y[c] - y[a]) -
         1LL * (y[b] - y[a]) * (x[c] - x[a]);
}
long long area[205][205];
int f[205][205];
int main(void) {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); ++(i)) scanf("%d %d", x + i, y + i);
  long long sum = 0;
  for (int(i) = 0; (i) < (n); ++(i)) sum += mul(0, (i + 1) % n, (i + 2) % n);
  if (sum < 0) {
    reverse(x, x + n);
    reverse(y, y + n);
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    f[i][(i + 1) % n] = 1;
  }
  for (int len = 2; len < n; ++len) {
    for (int i = 0; i < n; ++i) {
      int j = (i + len) % n;
      for (int k = (i + 1) % n; k != j; k = (k + 1) % n) {
        if (mul(i, k, j) > 0)
          f[i][j] = (f[i][j] + 1LL * f[i][k] * f[k][j]) % mod;
      }
    }
  }
  printf("%d\n", f[0][n - 1]);
  return 0;
}
