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 = 1000000000 + 7;
struct point {
  long long x, y;
  point() {}
  point(long long x, long long y) { this->x = x, this->y = y; }
  long long operator^(const point& p) { return x * p.y - p.x * y; }
  point operator-(const point& p) {
    point t;
    t.x = this->x - p.x, t.y = this->y - p.y;
    return t;
  }
};
point p[222];
long long dp[222][222];
int MAIN() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
  }
  long long area = 0;
  for (int i = 0; i < n; i++) {
    area += p[i] ^ p[(i + 1) % n];
  }
  if (area < 0) {
    reverse(p, p + n);
  }
  for (int l = 1; l < n; l++) {
    for (int i = 0; i + l < n; i++) {
      if (l == 1) {
        dp[i][i + l] = 1;
        continue;
      }
      int j = i + l;
      for (int k = i + 1; k < j; k++) {
        point a = p[k] - p[i], b = p[k] - p[j];
        if ((a ^ b) < 0) {
          dp[i][j] += (dp[i][k] * dp[k][j]) % MOD;
          dp[i][j] %= MOD;
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}
