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;
struct P {
  double x, y;
} a[259];
long long f[259][259];
int n;
int read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
double cross(P a, P b, P c) {
  P x = (P){b.x - a.x, b.y - a.y};
  P y = (P){c.x - a.x, c.y - a.y};
  return x.x * y.y - x.y * y.x;
}
bool jud() {
  double ans = 0;
  for (int i = 2; i <= n; i++) ans += cross(a[1], a[i - 1], a[i]);
  if (ans > 0) return 1;
  return 0;
}
void rever() {
  for (int i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].x, &a[i].y);
  if (jud()) rever();
  for (int i = 1; i <= n; i++) f[i][i + 1] = 1;
  for (int l = 1; l <= n - 1; l++)
    for (int i = 1; i + l <= n; i++)
      for (int j = i + 1; j < i + l; j++)
        f[i][i + l] =
            (f[i][i + l] + f[i][j] % 1000000007 * f[j][i + l] % 1000000007 *
                               (cross(a[i], a[j], a[i + l]) < 0 ? 1 : 0)) %
            1000000007;
  printf("%lld\n", f[1][n] % 1000000007);
  return 0;
}
