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 MODULUS = 1000000007;
long long tri(int x0, int y0, int x1, int y1, int x2, int y2) {
  long long a1 = x1 - x0;
  long long b1 = y1 - y0;
  long long a2 = x2 - x0;
  long long b2 = y2 - y0;
  return a1 * b2 - b1 * a2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> xs(n);
  vector<int> ys(n);
  for (int i = 0; i < n; ++i) cin >> xs[i] >> ys[i];
  long long total_area = 0;
  for (int i = 1; i < n - 1; ++i)
    total_area += tri(xs[0], ys[0], xs[i], ys[i], xs[i + 1], ys[i + 1]);
  if (total_area < 0) {
    reverse(xs.begin() + 1, xs.end());
    reverse(ys.begin() + 1, ys.end());
    total_area = -total_area;
  }
  vector<vector<long long> > area(n);
  for (int i = 0; i < n; ++i) area[i].assign(n, 0);
  vector<vector<int> > tab(n);
  for (int i = 0; i < n; ++i) {
    tab[i].assign(n, 0);
    tab[i][i + 1] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j)
      area[i][j] = area[i][j - 1] +
                   tri(xs[i], ys[i], xs[j - 1], ys[j - 1], xs[j], ys[j]);
  }
  for (int l = 2; l < n; ++l) {
    for (int i = 0; i < n - l; ++i) {
      if (area[i][i + l] <= 0) continue;
      for (int j = i + 1; j < i + l; ++j) {
        long long t = tri(xs[i], ys[i], xs[j], ys[j], xs[i + l], ys[i + l]);
        if (t > 0) {
          tab[i][i + l] =
              (int)((tab[i][i + l] + (long long)tab[i][j] * tab[j][i + l]) %
                    MODULUS);
        }
      }
    }
  }
  cout << tab[0][n - 1] << "\n";
  return 0;
}
