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 long long Mod = 1000000007;
long long f[5005][5005];
struct Point {
  double x, y;
  Point() {}
  Point(double x0, double y0) : x(x0), y(y0) {}
} p[500005];
Point operator-(Point p1, Point p2) { return Point(p1.x - p2.x, p1.y - p2.y); }
double operator*(Point p1, Point p2) { return p1.x * p2.y - p1.y * p2.x; }
int read() {
  int t = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    t = t * 10 + ch - '0';
    ch = getchar();
  }
  return t * f;
}
long long solve(int l, int r) {
  if (f[l][r] != -1) return f[l][r];
  if (l > r) return f[l][r] = 0;
  if (l + 1 == r) return f[l][r] = 1;
  f[l][r] = 0;
  for (int i = l + 1; i < r; i++) {
    if ((p[r] - p[l]) * (p[i] - p[l]) < 0)
      (f[l][r] += (solve(l, i) * solve(i, r)) % Mod) %= Mod;
  }
  return f[l][r];
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) p[i].x = read(), p[i].y = read();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) f[i][j] = -1;
  double res = 0;
  p[n + 1] = p[1];
  for (int i = 1; i <= n; i++) res += p[i] * p[i + 1];
  if (res < 0) {
    for (int i = 1; i <= n / 2; i++) std::swap(p[i], p[n - i + 1]);
  }
  printf("%I64d\n", solve(1, n));
}
