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 = 1e9 + 7;
long long dp[205][205], n;
int main() {
  cin >> n;
  vector<pair<long long, long long> > pt(n);
  for (long long i = 0; i < n; i++) {
    cin >> pt[i].first >> pt[i].second;
  }
  long long ori = 0;
  for (long long i = 0; i < n; i++) {
    ori += pt[i].first * pt[(i + 1) % n].second -
           pt[i].second * pt[(i + 1) % n].first;
  }
  if (ori > 0) {
    reverse(pt.begin(), pt.end());
  }
  for (long long len = 1; len < n; len++) {
    for (long long st = 0; st < n - len; st++) {
      long long l = st, r = st + len;
      if (len == 1)
        dp[l][r] = 1ll;
      else {
        dp[l][r] = 0;
        for (long long i = l + 1; i < r; i++) {
          long long a = pt[i].first - pt[l].first,
                    b = pt[i].second - pt[l].second;
          long long c = pt[i].first - pt[r].first,
                    d = pt[i].second - pt[r].second;
          if (a * d - b * c > 0) {
            dp[l][r] = (dp[l][r] + (dp[l][i] * dp[i][r]) % mod) % mod;
          }
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
