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 N = 205, P = 1000000007;
using ll = long long;
struct Point {
  int x, y;
};
Point operator+(const Point &a, const Point &b) {
  return {a.x + b.x, a.y + b.y};
}
Point operator-(const Point &a, const Point &b) {
  return {a.x - b.x, a.y - b.y};
}
ll cross(const Point &a, const Point &b) {
  return a.x * 1ll * b.y - a.y * 1ll * b.x;
}
int n;
ll dp[N][N];
Point a[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
  ll sum = cross(a[n], a[1]);
  for (int i = 1; i < n; i++) sum += cross(a[i], a[i + 1]);
  if (sum < 0) reverse(a + 1, a + n + 1);
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int l = 3; l <= n; l++)
    for (int i = 1; i + l - 1 <= n; i++) {
      int j = i + l - 1;
      for (int k = i + 1; k < j; k++)
        if (cross(a[k] - a[i], a[j] - a[k]) > 0)
          dp[i][j] = (dp[i][j] + dp[i][k] * dp[k][j] % P) % P;
    }
  cout << dp[1][n] << endl;
  return 0;
}
