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;
long long n, a, b, dp[300][300];
long long s;
vector<pair<long long, long long> > v;
long long r, tr;
long long gs(long long a, long long b, long long c) {
  long long s = 0;
  s = (v[b].first - v[a].first) * (v[c].second - v[a].second) -
      (v[b].second - v[a].second) * (v[c].first - v[a].first);
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  for (int i = 0; i < n; i++)
    s +=
        v[(i + 1) % n].first * v[i].second - v[(i + 1) % n].second * v[i].first;
  if (s < 0) reverse(v.begin(), v.end());
  for (int i = 0; i < n; i++) dp[i][(i + 1) % n] = 1;
  for (int span = 2; span < n; span++) {
    for (int l = 0; l < n; l++) {
      r = l + span;
      r %= n;
      for (int q = l + 1; q < l + span; q++) {
        tr = q % n;
        if (gs(l, r, tr) > 0)
          dp[l][r] = (dp[l][r] + dp[l][tr] * dp[tr][r]) % 1000000007;
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  cin.get();
  cin.get();
  return 0;
}
