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;
struct node {
  long long x, y;
  node(long long X = 0, long long Y = 0) { x = X, y = Y; }
};
node operator-(node a, node b) { return node(a.x - b.x, a.y - b.y); }
long long operator^(node a, node b) { return a.x * b.y - a.y * b.x; }
const int mo = 1000000007;
const int N = 205;
long long f[N][N];
node a[N];
int n;
void Init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld%lld", &a[i].x, &a[i].y);
}
void Changeit() {
  long long G = 0;
  for (int i = 2; i <= n; ++i) G += (a[i - 1] - a[1]) ^ (a[i] - a[1]);
  if (G < 0)
    for (int i = 1; i <= n >> 1; ++i) swap(a[i], a[n - i + 1]);
}
void Work() {
  Changeit();
  for (int i = 1; i <= n - 1; ++i) f[i][i + 1] = 1;
  for (int l = 2; l <= n - 1; ++l)
    for (int i = 1; i <= n - l; ++i) {
      int j = i + l;
      for (int k = i + 1; k <= j - 1; ++k)
        if (((a[k] - a[i]) ^ (a[j] - a[i])) > 0)
          (f[i][j] += f[i][k] * f[k][j]) %= mo;
    }
  printf("%lld\n", f[1][n]);
}
int main() {
  Init();
  Work();
  return 0;
}
