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 N = 205;
const int mod = 1e9 + 7;
int n, m;
long long f[N][N];
struct Vector {
  long long x, y;
} p[N];
Vector operator-(const Vector &a, const Vector &b) {
  return (Vector){a.x - b.x, a.y - b.y};
}
inline long long Cross(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }
inline long long Add(long long x, long long y) {
  return (x + y >= mod) ? (x + y - mod) : (x + y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  p[n + 1] = p[1];
  long long tmp = 0;
  for (int i = 1; i <= n; i++) tmp += Cross(p[i], p[i + 1]);
  if (tmp < 0) reverse(p + 1, p + n + 1);
  for (int i = 0; i < n; i++) {
    for (int j = 1; j + i <= n; j++) {
      if (i <= 1)
        f[j][i + j] = 1;
      else {
        for (int k = j + 1; k < i + j; k++) {
          if (Cross(p[k] - p[j], p[i + j] - p[j]) <= 0) continue;
          f[j][i + j] = Add(f[j][i + j], f[j][k] * f[k][i + j] % mod);
        }
      }
    }
  }
  cout << f[1][n];
  return 0;
}
