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 int MODP = 1000000007;
const int MAXN = 222;
long long n, x[MAXN], y[MAXN], sol[MAXN][MAXN];
long long ccw(int i, int j, int k) {
  return (x[j] - x[i]) * (y[k] - y[i]) - (x[k] - x[i]) * (y[j] - y[i]);
}
long long solve(int l, int r) {
  if (l == r - 1) return 1;
  if (sol[l][r] != -1) return sol[l][r];
  sol[l][r] = 0;
  for (int i = l + 1; i < r; i++)
    if (ccw(r, l, i) > 0)
      sol[l][r] = (sol[l][r] + solve(l, i) * solve(i, r)) % MODP;
  return sol[l][r];
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  memset(sol, -1, sizeof sol);
  long long area = 0;
  for (int i = 0; i < n; i++)
    area += x[i] * y[((i + 1) % n)] - x[((i + 1) % n)] * y[i];
  if (area < 0) {
    reverse(x, x + n);
    reverse(y, y + n);
  }
  cout << solve(0, n - 1) << endl;
}
