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 node {
  long long x, y;
  node() {}
  node(double _x, double _y) : x(_x), y(_y) {}
  void read() { scanf("%I64d%I64d", &x, &y); }
  long long operator%(const node &a) const { return x * a.y - a.x * y; }
  node operator-(const node &a) const { return node(x - a.x, y - a.y); }
} a[301], tmp[301];
int n;
long long sum, f[301][301];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) a[i].read();
  a[n + 1] = a[1];
  for (int i = 1; i <= n; i++) sum += a[i] % a[i + 1];
  if (sum < 0) {
    for (int i = 1; i <= n; i++) tmp[i] = a[n - i + 1];
    for (int i = 1; i <= n; i++) a[i] = tmp[i];
    a[n + 1] = a[1];
  }
  for (int i = 1; i < n; i++) f[i][i + 1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; i + j - 1 <= n; j++)
      for (int k = j + 1; k < i + j - 1; k++)
        if ((a[i + j - 1] - a[j]) % (a[i + j - 1] - a[k]) > 0)
          f[j][i + j - 1] =
              (f[j][i + j - 1] + f[j][k] * f[k][i + j - 1]) % 1000000007;
  printf("%I64d\n", f[1][n]);
}
