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 = 207;
const int P = 1e9 + 7;
struct Node {
  int x, y;
  Node operator-(const Node &p) const { return (Node){x - p.x, y - p.y}; }
  long long operator^(const Node &p) const {
    return 1ll * x * p.y - 1ll * y * p.x;
  }
} a[N];
int n, dp[N][N];
inline void inc(int &x, int y) {
  if ((x += y) >= P) x -= P;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &a[i].x, &a[i].y);
  long long area = 0;
  for (int i = (0); i < (n); ++i) area += a[i] ^ a[(i + 1) % n];
  if (area > 0) reverse(a, a + n);
  for (int i = (0); i < (n - 1); ++i) dp[i][i + 1] = 1;
  for (int d = (2); d < (n); ++d)
    for (int i = (0); i < (n - d); ++i) {
      int j = i + d;
      for (int k = (i + 1); k < (j); ++k)
        if (((a[k] - a[i]) ^ (a[k] - a[j])) > 0)
          inc(dp[i][j], 1ll * dp[i][k] * dp[k][j] % P);
    }
  printf("%d\n", dp[0][n - 1]);
  return 0;
}
