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 N = 210, B = 1e9 + 7;
int n;
long long x[N], y[N], d[N][N];
bool fl[N][N];
long long cross(long long x, long long y, long long xx, long long yy) {
  return x * yy - xx * y;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld%lld", x + i, y + i);
  x[n] = x[0];
  y[n] = y[0];
  long long area = 0;
  for (int k = 0; k < n; ++k) area += cross(x[k], y[k], x[k + 1], y[k + 1]);
  if (area < 0) {
    reverse(x, x + n + 1);
    reverse(y, y + n + 1);
  }
  for (int i = 0; i < n; ++i) {
    d[i][i + 1] = 1;
  }
  for (int k = 2; k < n; ++k) {
    for (int i = 0; i + k <= n; ++i) {
      int b = i + k;
      for (int j = i; j <= b; ++j) {
        if (cross(x[i] - x[j], y[i] - y[j], x[b] - x[j], y[b] - y[j]) < 0)
          d[i][b] = (d[i][b] + d[i][j] * d[j][b] % B) % B;
      }
    }
  }
  printf("%lld\n", d[0][n - 1]);
  return 0;
}
