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 mod = 1000000007;
struct Point {
  double x, y;
  Point operator-(const Point &B) const { return (Point){x - B.x, y - B.y}; }
  double operator^(const Point &B) const { return x * B.y - y * B.x; }
} p[205];
int f[205][205];
int dfs(int l, int r) {
  if (l + 1 == r) return 1;
  if (f[l][r] != -1) return f[l][r];
  int ans = 0;
  for (int k = l + 1; k < r; k++)
    if (((p[k] - p[l]) ^ (p[r] - p[l])) > 1e-10)
      ans = (ans + 1ll * dfs(l, k) * dfs(k, r)) % mod;
  return f[l][r] = ans;
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) f[i][j] = -1;
  for (int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
  double ans = p[n] ^ p[1];
  for (int i = 1; i < n; i++) ans += p[i] ^ p[i + 1];
  if (ans < 0.0) reverse(p + 1, p + n + 1);
  printf("%d", dfs(1, n));
  return 0;
}
