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 mo = 1e9 + 7;
int n, x[205], y[205], dp[205][205];
void ntr(int &x) { x = x >= mo ? x - mo : (x < 0 ? x + mo : x); }
void inc(int &x, int y) { x += y, ntr(x); }
long long cross(int x1, int y1, int x2, int y2) {
  return 1ll * x1 * y2 - 1ll * x2 * y1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", x + i, y + i);
  x[n + 1] = x[1], y[n + 1] = y[1];
  long long s = 0;
  for (int i = 1; i <= n; i++) s += cross(x[i], y[i], x[i + 1], y[i + 1]);
  if (s > 0) {
    for (int i = 1; i <= n / 2; i++)
      swap(x[i], x[n + 1 - i]), swap(y[i], y[n + 1 - i]);
  }
  for (int i = 1; i <= n - 1; i++) dp[i][i + 1] = 1;
  for (int i = 2; i <= n - 1; i++)
    for (int j = 1; j <= n - i; j++)
      for (int k = j + 1; k <= j + i - 1; k++)
        if (cross(x[j + i] - x[j], y[j + i] - y[j], x[k] - x[j], y[k] - y[j]) >
            0)
          inc(dp[j][j + i], 1ll * dp[j][k] * dp[k][j + i] % mo);
  printf("%d\n", dp[1][n]);
  return 0;
}
