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 long long mod = 1000000007;
long long n, f[300][300], sum;
bool vis[300][300];
struct node {
  long long x, y;
  node(long long x = 0, long long y = 0) : x(x), y(y) {}
  node operator-(const node &p) const { return node(x - p.x, y - p.y); }
} a[300];
long long cross(node u, node v) { return u.x * v.y - u.y * v.x; }
long long dp(long long l, long long r) {
  if (r - l == 1) return 1;
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  for (long long k = l + 1; k < r; ++k) {
    if (cross(a[r] - a[k], a[l] - a[k]) * sum > 0) {
      f[l][r] = (f[l][r] + dp(l, k) * dp(k, r)) % mod;
    }
  }
  return f[l][r];
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) {
    scanf("%lld%lld", &a[i].x, &a[i].y);
  }
  for (long long i = 1; i < n; ++i) {
    sum += cross(a[i] - a[n], a[i + 1] - a[n]);
  }
  sum = sum > 0 ? 1 : -1;
  printf("%lld\n", dp(1, n));
  return 0;
}
