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 int N = 200 + 5;
const int Mod = 1e9 + 7;
inline int Get() {
  char ch;
  int Num, Sign = 0;
  while ((ch = getchar()) < '0' || ch > '9')
    if (ch == '-') Sign ^= 1;
  Num = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9')
    Num = (Num << 3) + (Num << 1) + ch - '0';
  return Sign ? -Num : Num;
}
int n, x[N], y[N], g[N][N];
double sum;
struct node {
  double x, y;
  inline node(){};
  inline node(double a, double b) { x = a, y = b; }
  inline double friend operator*(const node &a, const node &b) {
    return a.x * b.y - b.x * a.y;
  }
  inline node friend operator-(const node &a, const node &b) {
    return node(a.x - b.x, a.y - b.y);
  }
} a[N];
int main() {
  n = Get();
  for (int i = 0; i < n; ++i) a[i].x = Get(), a[i].y = Get();
  for (int i = 0; i < n; ++i) sum += a[i] * a[(i + 1) % n];
  if (sum < 0)
    for (int i = 0; i * 2 < n; ++i) std ::swap(a[i], a[n - i - 1]);
  for (int len = 1; len < n; ++len)
    for (int l = 0; l < n; ++l) {
      int r = l + len;
      if (r >= n) break;
      if (len == 1) g[l][r] = 1;
      for (int k = l + 1; k < r; ++k)
        if ((a[l] - a[r]) * (a[k] - a[r]) > 0)
          (g[l][r] += (long long)g[l][k] * g[k][r] % Mod) %= Mod;
    }
  printf("%d\n", g[0][n - 1]);
}
