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;
const int P = 1000000007;
int n, x[205], y[205], f[205][205];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", x + i, y + i);
  long long tmp = 0;
  for (int i = 0; i < n; i++)
    tmp += 1ll * x[i] * y[(i + 1) % n] - 1ll * y[i] * x[(i + 1) % n];
  if (tmp < 0) reverse(x, x + n), reverse(y, y + n);
  for (int i = 0; i < n - 1; i++) f[i][i + 1] = 1;
  for (int i = n - 3; i >= 0; i--)
    for (int j = i + 2; j < n; j++)
      for (int k = i + 1; k < j; k++)
        if (1ll * (x[i] - x[k]) * (y[j] - y[k]) <
            1ll * (y[i] - y[k]) * (x[j] - x[k]))
          f[i][j] = (f[i][j] + 1ll * f[i][k] * f[k][j]) % P;
  printf("%d\n", f[0][n - 1]);
  return 0;
}
