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 maxn = 1100;
const int mod = 1e9 + 7;
struct point {
  long long x, y;
  point() {}
  point(long long x, long long y) : x(x), y(y) {}
};
point a[maxn];
int f[maxn][maxn];
long long det(const point& a, const point& b) { return a.x * b.y - a.y * b.x; }
point operator-(const point& a, const point& b) {
  return point(a.x - b.x, a.y - b.y);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y);
  a[n + 1] = a[1];
  long long tmp = 0;
  for (int i = 1; i <= n; i++) {
    tmp += det(a[i], a[i + 1]);
  }
  if (tmp < 0) reverse(a + 1, a + n + 1);
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n - 1; i++) f[i][i + 1] = 1;
  for (int len = 3; len <= n; len++) {
    for (int i = 1; i <= n; i++) {
      int j = i + len - 1;
      if (j > n) break;
      for (int k = i + 1; k < j; k++) {
        if (det(a[j] - a[i], a[k] - a[i]) < 0)
          f[i][j] += 1LL * f[i][k] * f[k][j] % mod;
        f[i][j] %= mod;
      }
    }
  }
  printf("%d\n", f[1][n]);
}
