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 N = 210, kmod = 1000000007;
const double eps = 1e-8;
inline int sig(double x) { return (x > eps) - (x < -eps); }
int dp[N][N], n;
struct Point {
  double x, y;
  void in() { scanf("%lf%lf", &x, &y); }
  Point(double x = 0., double y = 0.) : x(x), y(y) {}
  Point operator-(const Point &o) const { return Point(x - o.x, y - o.y); }
  double operator^(const Point &o) const { return x * o.y - y * o.x; }
  double Cross(Point &a, Point &b) { return (a - *this) ^ (b - *this); }
};
vector<Point> v;
double getArea() {
  if (n < 3) return 0.;
  double s = v[0].y * (v[n - 1].x - v[1].x);
  for (int i = 1; i < n - 1; ++i) s += v[i].y * (v[i - 1].x - v[i + 1].x);
  s += v[n - 1].y * (v[n - 2].x - v[0].x);
  return s;
}
int main() {
  Point o;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) o.in(), v.push_back(o), dp[i][i + 1] = 1;
  int q = 0;
  if (sig(getArea()) < 0) reverse(v.begin(), v.end());
  for (int l = 2; l < n; ++l)
    for (int i = 0; i < n - l; ++i) {
      int j = i + l;
      for (int k = i + 1; k < j; ++k) {
        if (sig(v[i].Cross(v[k], v[j])) == 1)
          dp[i][j] =
              (dp[i][j] + ((long long)dp[i][k] * dp[k][j]) % kmod) % kmod;
      }
    }
  printf("%d\n", dp[0][n - 1]);
}
