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 jsb = 1000000007;
const int N = 205;
inline long long cha(int x1, int y1, int x2, int y2) {
  return x1 * 1ll * y2 - x2 * 1ll * y1;
}
int f[N][N];
bool vis[N][N];
int n;
int x[N], y[N];
int dp(int l, int r) {
  if (vis[l][r]) return f[l][r];
  if (l == r - 1) return 1;
  vis[l][r] = 1;
  for (int k = (l + 1); k <= (int)r - 1; k++)
    if (cha(x[k] - x[l], y[k] - y[l], x[r] - x[l], y[r] - y[l]) > 0)
      f[l][r] = (f[l][r] + dp(l, k) * 1ll * dp(k, r)) % jsb;
  return f[l][r];
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)n; i++) scanf("%d%d", &x[i], &y[i]);
  x[n + 1] = x[1];
  y[n + 1] = y[1];
  long long area = 0;
  for (int i = (1); i <= (int)n; i++)
    area += cha(x[i], y[i], x[i + 1], y[i + 1]);
  if (area < 0) {
    for (int i = (1); i <= (int)n / 2; i++) {
      swap(x[i], x[n + 1 - i]);
      swap(y[i], y[n + 1 - i]);
    }
    x[n + 1] = x[1];
    y[n + 1] = y[1];
  }
  printf("%d\n", dp(1, n));
  return 0;
}
