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;
pair<long long, long long> p[403 + 5];
bool OK[403][403];
long long f[403][403];
long long isleft(pair<long long, long long> a, pair<long long, long long> b,
                 pair<long long, long long> c) {
  return (a.first * b.second - a.second * b.first) +
         (b.first * c.second - b.second * c.first) +
         (c.first * a.second - c.second * a.first);
}
int main() {
  int i, j, k, l;
  long long x, y;
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  long long area = 0LL;
  for (i = 1; i <= n; i++) {
    cin >> x >> y;
    p[i] = pair<long long, long long>(x, y);
  }
  p[0] = pair<long long, long long>(0LL, 0LL);
  p[n + 1] = p[1];
  for (i = 2; i <= n + 1; i++) {
    area += isleft(p[0], p[i - 1], p[i]);
  }
  if (area < 0) {
    reverse(p + 1, p + n + 1);
  }
  for (i = 1; i <= n; i++) f[i][i + 1] = 1LL;
  for (l = 2; l <= n; l++) {
    for (i = 1; i <= n; i++) {
      j = (i + l);
      if (j > n) break;
      f[i][j] = 0LL;
      for (k = i + 1; k < j; k++) {
        if (isleft(p[i], p[k], p[j]) <= 0) continue;
        f[i][j] = (f[i][j] + (f[i][k] * f[k][j]) % 1000000007 + 1000000007) %
                  1000000007;
      }
    }
  }
  cout << (f[1][n] + 1000000007) % 1000000007 << "\n";
  return 0;
}
