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 infi = 2147483647;
const int mod = 1000000007;
const double eps = 1e-6;
const int N = 250;
struct node {
  int x, y;
  void read() { scanf("%d%d", &x, &y); }
  node(int _x = 0, int _y = 0) {
    x = _x;
    y = _y;
  }
} v[N];
int n, f[N][N];
long long S;
bool cross(node a, node b) {
  return (long long)a.x * b.y - (long long)a.y * b.x > 0;
}
bool isok(node a, node b, node c) {
  return cross(node(a.x - b.x, a.y - b.y), node(a.x - c.x, a.y - c.y));
}
long long area(node a, node b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
int main() {
  int i, j, k, d;
  scanf("%d", &n);
  for (i = 0; i < n; i++) v[i].read();
  for (i = 0; i < n; i++) S += area(v[i], v[(i + 1) % n]);
  if (S > 0)
    for (i = 0; i < n / 2; i++) swap(v[i], v[n - i - 1]);
  for (i = 0; i < n - 1; i++) f[i][i + 1] = 1;
  for (d = 2; d < n; d++)
    for (i = 0; i < n - d; i++) {
      j = i + d;
      for (k = i + 1; k < j; k++)
        if (isok(v[k], v[i], v[j]))
          f[i][j] = (f[i][j] + (long long)f[i][k] * f[k][j]) % mod;
    }
  printf("%d\n", f[0][n - 1]);
  return 0;
}
