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;
typedef long long int ll;
typedef vector<int> vi;
const long long mod = 1e9 + 7;
struct st {
  double x, y;
} a[201];
int n;
long long f[201][201];
double s;
st operator-(st a, st b) {
  st c;
  c.x = a.x - b.x;
  c.y = a.y - b.y;
  return c;
}
double operator*(st a, st b) { return a.x * b.y - a.y * b.x; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].x >> a[i].y;
    if (i > 1) s += a[i - 1] * a[i];
  }
  s += a[n] * a[1];
  if (s > 0)
    for (int i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
  for (int i = 1; i < n; i++) f[i][i + 1] = 1;
  for (int l = 3; l <= n; l++) {
    for (int i = 1; i <= n - l + 1; i++) {
      int j = i + l - 1;
      for (int k = i + 1; k < j; k++)
        if ((a[j] - a[i]) * (a[k] - a[i]) > 0)
          f[i][j] = (f[i][j] + f[i][k] * f[k][j]) % mod;
    }
  }
  cout << f[1][n];
}
