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;
struct point {
  long long x, y;
  point(long long X = 0, long long Y = 0) { x = X, y = Y; }
} p[310];
point V(point s, point e) { return point(e.x - s.x, e.y - s.y); }
long long crossP(point a, point b) { return a.x * b.y - a.y * b.x; }
const long long mod = 1000000007;
bool vis[310][310];
long long f[310][310];
long long dp(int l, int r) {
  long long &x = f[l][r];
  if (vis[l][r]) return x;
  if (l + 1 == r) return x = 1;
  vis[l][r] = true;
  for (int i = l + 1; i <= r - 1; i++) {
    if (crossP(V(p[r], p[l]), V(p[r], p[i])) < 0)
      x = (x + dp(l, i) * dp(i, r)) % mod;
  }
  return x;
}
int main(int argc, char *argv[]) {
  cin >> n;
  long long area = 0;
  for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y;
  p[n + 1] = p[1];
  for (int i = 1; i <= n; i++) area += crossP(p[i], p[i + 1]);
  if (area > 0)
    for (int i = 1; 2 * i <= n; i++) swap(p[i], p[n - i + 1]), p[n + 1] = p[1];
  cout << dp(1, n) << endl;
  return 0;
}
