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 = 205;
const int mod = 1000000007;
int a, b, c, d, e, f, g, h, i, j, k;
struct Point {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  friend Point operator-(const Point &x, const Point &y) {
    return Point(x.x - y.x, x.y - y.y);
  }
  friend long long operator^(const Point &x, const Point &y) {
    return (long long)x.x * y.y - (long long)x.y * y.x;
  }
};
Point G[N];
int F[N][N];
void Init() {
  cin >> a;
  for (int ii = 0; ii < a; ii++) scanf("%d%d", &G[ii].x, &G[ii].y);
  long long S = 0;
  for (int ii = 0; ii < a; ii++) S += (G[ii] ^ G[(ii + 1) % a]);
  if (S < 0) reverse(G, G + a);
}
void Work() {
  for (int dd = 1; dd < a; dd++) {
    for (int ii = 0, jj = ii + dd; jj < a; ii++, jj++) {
      if (dd == 1) {
        F[ii][jj] = 1;
        continue;
      }
      for (int kk = ii + 1; kk < jj; kk++)
        if (((G[kk] - G[ii]) ^ (G[kk] - G[jj])) < 0)
          (F[ii][jj] += (long long)F[ii][kk] * F[kk][jj] % mod) %= mod;
    }
  }
  cout << F[0][a - 1] << endl;
}
int main() {
  Init(), Work();
  return 0;
}
