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 = 300 + 5;
const long long MOD = 1e9 + 7;
struct Point {
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  void read() { cin >> x >> y; }
  Point operator-(const Point &t) const { return Point(x - t.x, y - t.y); }
  double operator^(const Point &t) const { return x * t.y - y * t.x; }
} A[N];
int n;
long long f[N][N];
double g[N][N];
double Cross(Point A, Point B, Point C) { return ((B - A) ^ (C - A)); }
void Change(Point *A) {
  for (int i = 1; i <= n / 2; i++) swap(A[i], A[n - i + 1]);
}
bool Check(Point *A) {
  double area = 0;
  for (int i = 1; i <= n; i++) area += A[i] ^ A[i % n + 1];
  return area > 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) A[i].read();
  if (Check(A)) Change(A);
  for (int i = 1; i <= n; i++) f[i][i + 1] = 1ll, g[i][i + 1] = 0;
  for (int i = n; i >= 1; i--)
    for (int j = i; j <= n; j++)
      for (int k = i + 1; k <= j - 1; k++) {
        double s = Cross(A[i], A[j], A[k]) / 2;
        if (s <= 0)
          continue;
        else
          s = s * s;
        f[i][j] = (f[i][j] + f[i][k] * f[k][j] % MOD) % MOD;
      }
  cout << f[1][n] << endl;
  return 0;
}
