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 maxn = 200 + 10;
const int mod = 1e9 + 7;
typedef struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
  Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
  long long operator*(const Point& p) { return x * p.y - y * p.x; }
} P;
P p[maxn];
long long dp[maxn][maxn];
int main(void) {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  long long d = 0;
  for (int i = 0; i < n; i++) d += p[i] * p[(i + 1) % n];
  if (d < 0) reverse(p, p + n);
  for (int i = 0; i < n; i++) dp[i][i + 1] = 1;
  for (int len = 2; len < n; len++) {
    for (int l = 0; l + len < n; l++) {
      int r = l + len;
      for (int k = l + 1; k < r; k++) {
        if ((p[k] - p[l]) * (p[l] - p[r]) < 0)
          dp[l][r] = (dp[l][r] + (dp[l][k] * dp[k][r]) % mod) % mod;
        else
          continue;
      }
    }
  }
  cout << dp[0][n - 1] << endl;
}
