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 maxn = 201;
const long long mo = 1e9 + 7;
int n;
struct vt {
  long long x, y;
  void read() { scanf("%lld%lld", &x, &y); }
  vt operator-(const vt &k) const { return {x - k.x, y - k.y}; }
  long long cross(const vt &k) const { return x * k.y - y * k.x; }
} p[maxn];
long long dp[maxn][maxn];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    p[i].read();
  }
  long long area = 0;
  for (int i = 0; i < n; ++i) {
    area += p[i].cross(p[(i + 1) % n]);
  }
  if (area < 0) {
    std::reverse(p, p + n);
  }
  for (int i = 0; i + 1 < n; ++i) {
    dp[i][i + 1] = 1;
  }
  for (int i = 2; i < n; ++i) {
    for (int j = 0; i + j < n; ++j) {
      for (int k = j + 1; k < i + j; ++k) {
        if ((p[k] - p[j]).cross(p[i + j] - p[k]) > 0) {
          dp[j][i + j] = (dp[j][i + j] + dp[j][k] * dp[k][i + j]) % mo;
        }
      }
    }
  }
  printf("%lld\n", dp[0][n - 1]);
  return 0;
}
