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 s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
  return s * f;
}
const int N = 205, mod = 1e9 + 7;
inline int add(int a, int b) { return (a + b) >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
struct Node {
  int x, y;
};
inline Node operator-(Node a, Node b) { return (Node){a.x - b.x, a.y - b.y}; }
inline long long operator*(Node a, Node b) {
  return 1ll * a.x * b.y - 1ll * a.y * b.x;
}
Node a[N];
int n;
bool vis[N][N];
int f[N][N];
int dp(int l, int r) {
  if (r == l + 1) return 1;
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  f[l][r] = 0;
  for (int i = l + 1; i < r; ++i) {
    if ((a[i] - a[l]) * (a[r] - a[l]) > 0)
      f[l][r] = add(f[l][r], mul(dp(l, i), dp(i, r)));
  }
  return f[l][r];
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    a[i].x = read();
    a[i].y = read();
  }
  long long S = 0;
  for (int i = 2; i < n; ++i) S += (a[i] - a[1]) * (a[i + 1] - a[1]);
  if (S < 0) reverse(a + 1, a + 1 + n);
  printf("%d", dp(1, n));
  return 0;
}
