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 N, F[205][205];
struct Vec {
  int x, y;
  Vec operator-(const Vec& b) { return (Vec){x - b.x, y - b.y}; }
  long long operator*(const Vec& b) { return 1ll * x * b.y - 1ll * y * b.x; }
} A[205];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d%d", &A[i].x, &A[i].y);
  A[0] = A[N];
  long long S = 0;
  for (int i = 1; i <= N; i++) S += A[i] * A[i - 1];
  if (S >= 0)
    for (int i = 1; i <= N / 2; i++) swap(A[i], A[N - i + 1]);
  for (int i = 1; i <= N - 1; i++) F[i][i + 1] = 1;
  for (int l = 3; l <= N; l++)
    for (int i = 1; i <= N - l + 1; i++) {
      int j = i + l - 1;
      for (int k = i + 1; k <= j - 1; k++)
        if ((A[k] - A[i]) * (A[j] - A[i]) > 0)
          F[i][j] =
              (1ll * F[i][k] * F[k][j] % 1000000007 + F[i][j]) % 1000000007;
    }
  printf("%d\n", F[1][N]);
}
