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>
static const long long N = 205;
static const long long MOD = 1000000007;
long long f[N][N];
struct Point {
  long long x, y;
  Point(const long long &_x = 0, const long long &_y = 0) : x(_x), y(_y) {}
  Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }
  long long operator*(const Point &p) const { return x * p.y - y * p.x; }
} p[N];
void read_n(long long &n) { scanf("%lld", &n); }
void read_p(Point &p) { scanf("%lld%lld", &p.x, &p.y); }
signed main() {
  long long n, sum = 0;
  read_n(n), std ::for_each(p, p + n, read_p);
  for (long long i = 1; i < n - 1; ++i)
    sum += (p[i] - p[0]) * (p[i + 1] - p[0]);
  if (sum < 0) std ::reverse(p, p + n);
  for (long long i = 0; i < n; ++i) f[i][(i + 1) % n] = 1;
  for (long long i = 2; i <= n; ++i)
    for (long long j = 0; j < n; ++j) {
      long long k = (j + i) % n;
      for (long long l = (j + 1) % n; l != k; (++l) %= n)
        if ((p[l] - p[j]) * (p[k] - p[j]) > 0)
          (f[j][k] += f[j][l] * f[l][k]) %= MOD;
    }
  printf("%lld\n", f[0][n - 1]);
  return 0;
}
