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>
const double pi = acos(-1.0);
const double eps = 1e-9;
using namespace std;
const int N = 333;
const long long MOD = 1e9 + 7;
long long dp[N][N];
vector<pair<long long, long long>> v;
int n;
long long triangle_area_2(long long x1, long long y1, long long x2,
                          long long y2, long long x3, long long y3) {
  return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v.push_back({x, y});
  }
  long long s = 0;
  for (int i = 0; i < n; i++) {
    int id = (i + 1) % n;
    s += v[id].first * v[i].second - v[id].second * v[i].first;
    dp[i][id] = 1;
  }
  if (s < 0) {
    reverse((v).begin(), (v).end());
  }
  for (int len = 2; len < n; len++) {
    for (int l = 0; l < n; l++) {
      int r = (l + len) % n;
      for (int take = l + 1; take < l + len; take++) {
        int id = take % n;
        if (triangle_area_2(v[l].first, v[l].second, v[id].first, v[id].second,
                            v[r].first, v[r].second) < 0) {
          dp[l][r] = (dp[l][r] + dp[l][id] * dp[id][r]) % MOD;
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(3);
  int step = 1;
  for (int i = 0; i < step; i++) {
    solve();
  }
}
