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;
struct point {
  long long int x, y;
  long long int operator*(const point& c) const { return x * c.y - y * c.x; }
  point operator-(const point& c) const { return (point){x - c.x, y - c.y}; }
} p[205];
int n;
long long int dp[205][205];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &p[i].x, &p[i].y);
  }
  long long int tmp = p[n] * p[1];
  for (int i = 1; i < n; i++) tmp += p[i] * p[i + 1];
  if (tmp < 0) {
    reverse(p + 1, p + n + 1);
  }
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int k = 2; k < n; k++)
    for (int i = 1; i + k <= n; i++)
      for (int j = 1; j < k; j++) {
        if ((p[i + j] - p[i]) * (p[i + k] - p[i]) > 0) {
          dp[i][i + k] =
              (dp[i][i + k] + (dp[i][i + j] * dp[i + j][i + k])) % 1000000007;
        }
      }
  printf("%lld\n", dp[1][n]);
  return 0;
}
