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;
inline int read() {
  int w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 305, mod = 1e9 + 7;
int n, f[maxn][maxn];
struct vec {
  int x, y;
  vec(int x = 0, int y = 0) : x(x), y(y) {}
  vec operator+(const vec &a) const { return vec(x + a.x, y + a.y); }
  vec operator-(const vec &a) const { return vec(x - a.x, y - a.y); }
  long long operator*(const vec &a) const {
    return 1ll * x * a.y - 1ll * y * a.x;
  }
} a[maxn];
inline void work() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
  long long res = a[n] * a[1];
  for (int i = 1; i < n; i++) res += a[i] * a[i + 1], f[i][i + 1] = 1;
  if (res < 0) reverse(a + 1, a + 1 + n);
  for (int len = 2; len < n; len++)
    for (int l = 1, r = l + len; r <= n; l++, r++)
      for (int k = l + 1; k < r; k++)
        if ((a[r] - a[l]) * (a[k] - a[l]) < 0)
          f[l][r] = (f[l][r] + 1ll * f[l][k] * f[k][r]) % mod;
  printf("%d\n", f[1][n]);
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}
