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;
int A[205], B[205], i, j, k, n;
long long s, dp[205][205];
long long work(long long a, long long b, long long c, long long d) {
  return a * d - b * c;
}
const int MOD = 1000000007;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d%d", &A[i], &B[i]);
  A[n + 1] = A[1];
  B[n + 1] = B[1];
  for (i = 1; i <= n; i++) s += work(A[i], B[i], A[i + 1], B[i + 1]);
  if (s > 0)
    for (i = 1; i <= n / 2; i++) {
      swap(A[i], A[n - i + 1]);
      swap(B[i], B[n - i + 1]);
    }
  for (i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (i = 2; i < n; i++)
    for (j = 1; j <= n - i; j++)
      for (k = j + 1; k < i + j; k++)
        if (work(A[i + j] - A[j], B[i + j] - B[j], A[k] - A[j], B[k] - B[j]) >
            0)
          dp[j][i + j] = (dp[j][i + j] + dp[j][k] * dp[k][i + j]) % MOD;
  cout << dp[1][n];
  return 0;
}
