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;
double s;
long long f[1010][1010];
struct nodes {
  double x, y;
  nodes friend operator-(const nodes &a, const nodes &b) {
    return (nodes){a.x - b.x, a.y - b.y};
  }
  double friend operator*(const nodes &a, const nodes &b) {
    return a.x * b.y - a.y * b.x;
  }
} base[1010];
int main() {
  cin >> N;
  int i, j, k;
  for (i = 1; i <= N; ++i) scanf("%lf%lf", &base[i].x, &base[i].y);
  s += base[N] * base[1];
  for (i = 1; i < N; ++i) s += base[i] * base[i + 1], f[i][i + 1] = 1;
  if (s > 0) reverse(base + 1, base + N + 1);
  for (i = 1; i <= N; ++i)
    for (j = i - 2; j >= 1; --j)
      for (k = j; k <= i; ++k)
        if ((base[i] - base[j]) * (base[k] - base[j]) > 0)
          (f[j][i] += f[j][k] * f[k][i]) %= 1000000007;
  cout << f[1][N] << endl;
}
