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;
long long i, j, k, l, n, m, s, sum, A, C, tot, an;
const long long pp = 1000000007;
const long long N = 210;
class arr {
 public:
  long long x, y;
} a[N];
long long f[N][N];
long long X(long long x1, long long y1, long long x2, long long y2) {
  return (x1 * y2 - x2 * y1);
}
int main() {
  scanf("%I64d", &n);
  for (long long i = 1; i <= n; i++) scanf("%I64d%I64d", &a[i].x, &a[i].y);
  a[n + 1] = a[1];
  for (long long i = 1; i <= n; i++)
    s += X(a[i].x, a[i].y, a[i + 1].x, a[i + 1].y);
  if (s > 0)
    for (long long i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
  for (long long i = 1; i <= n; i++) f[i][i + 1] = 1;
  for (long long l = 3; l <= n; l++)
    for (long long i = 1; i + l - 1 <= n; i++) {
      long long j = i + l - 1;
      for (long long k = i + 1; k < j; k++)
        if (X(a[k].x - a[i].x, a[k].y - a[i].y, a[k].x - a[j].x,
              a[k].y - a[j].y) > 0)
          f[i][j] += f[i][k] * f[k][j] % pp, f[i][j] %= pp;
    }
  printf("%I64d\n", f[1][n]);
  return 0;
}
