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;
using point = complex<long long>;
const int N = 200, M = 1000000007;
int ways[N][N];
point p[N];
long long cross(point a, point b) {
  return a.real() * b.imag() - a.imag() * b.real();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    p[i] = point(x, y);
  }
  long long area = 0;
  for (int i = 0; i < n; ++i) {
    area += cross(p[i], p[(i + 1) % n]);
  }
  if (area < 0) {
    reverse(p, p + n);
  }
  for (int i = n - 2; i >= 0; --i) {
    ways[i][i + 1] = 1;
    for (int j = i + 2; j < n; ++j) {
      for (int k = i + 1; k < j; ++k) {
        if (cross(p[i], p[k]) + cross(p[k], p[j]) + cross(p[j], p[i]) > 0) {
          ways[i][j] = (ways[i][j] + (long long)ways[i][k] * ways[k][j]) % M;
        }
      }
    }
  }
  cout << ways[0][n - 1] << "\n";
}
