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 = 500;
const int mod = 1e9 + 7;
struct point {
  double x, y;
  double operator*(const point &a) const { return x * a.y - a.x * y; }
  point operator-(const point &a) const { return point{x - a.x, y - a.y}; }
} p[maxn];
long long dp[maxn][maxn];
int n;
int main() {
  cin >> n;
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < n; i++) cin >> p[i].x >> p[i].y;
  p[n] = p[0];
  double sum = 0;
  for (int i = 0; i < n; i++) sum += p[i] * p[i + 1];
  if (sum < 0) reverse(p, p + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) dp[i][i + 1] = 1;
  for (int len = 2; len < n; len++) {
    for (int i = 0; i < n; i++) {
      int j = i + len;
      if (j >= n) break;
      for (int k = i + 1; k < j; k++) {
        if ((p[i] - p[j]) * (p[k] - p[j]) > 0) {
          dp[i][j] = dp[i][j] + dp[i][k] * dp[k][j] % mod;
          dp[i][j] %= mod;
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
