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;
int n;
long long dp[210][210];
struct point {
  long long x, y;
  void read() { scanf("%I64d%I64d", &x, &y); }
} P[210];
long long cross(point a, point b, point c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
int main() {
  long long s = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) P[i].read();
  P[n + 1] = P[1];
  for (int i = 2; i <= n; i++) s += cross(P[1], P[i], P[i + 1]);
  if (s > 0)
    for (int i = 1; i <= n / 2; i++) swap(P[i], P[n - i + 1]);
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int i = 2; i < n; i++)
    for (int j = 1; j <= n - i; j++)
      for (int k = j + 1; k < i + j; k++)
        if (cross(P[j], P[i + j], P[k]) > 0)
          dp[j][i + j] = (dp[j][i + j] + dp[j][k] * dp[k][i + j]) % 1000000007;
  printf("%I64d", dp[1][n]);
}
