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 long long mod = 1000000007;
struct point {
  long long x, y;
} p[300];
long long cross(const point &a, const point &b) {
  return a.x * b.y - a.y * b.x;
}
long long f[300][300];
int main() {
  int T = 1;
  while (T--) {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i += 1) scanf("%lld %lld", &p[i].x, &p[i].y);
    long long rot = 0;
    for (int i = 0; i < n; i += 1) rot += cross(p[i], p[(i + 1) % n]);
    if (rot < 0) reverse(p, p + n);
    for (int i = 0; i < n; i += 1) fill(f[i], f[i] + n, 0LL);
    for (int i = 0; i + 1 < n; i += 1) f[i][i + 1] = 1;
    for (int i = n - 3; i >= 0; i -= 1)
      for (int j = i + 2; j < n; j += 1)
        for (int k = i + 1; k < j; k += 1)
          if (cross(p[i], p[j]) + cross(p[j], p[k]) < cross(p[i], p[k]))
            f[i][j] = (f[i][j] + f[i][k] * f[k][j]) % mod;
    printf("%lld\n", f[0][n - 1]);
  }
}
