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;
struct point {
  double first, second;
  point() {}
  point(double _x, double _y) {
    first = _x;
    second = _y;
  }
  point operator-(const point &a) {
    return point(first - a.first, second - a.second);
  }
  double operator*(const point &a) {
    return first * a.second - second * a.first;
  }
};
long mod = 1000 * 1000 * 1000 + 7, n;
long long d[301][301];
point v[301];
void init() {
  double tmp = 0;
  v[n] = v[0];
  for (int i = 0; i < n; i++) tmp += v[i] * v[i + 1];
  if (tmp < 0) reverse(v, v + n);
}
long long solve(int l, int r) {
  if (d[l][r] != -1) return d[l][r];
  long long ans = 0;
  if (r - l == 1) return 1;
  for (int i = l + 1; i < r; i++) {
    if ((v[l] - v[r]) * (v[i] - v[r]) > 0) {
      ans = (ans + solve(l, i) * solve(i, r)) % mod;
    }
  }
  return d[l][r] = ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    point t = point(1, 1);
    cin >> t.first >> t.second;
    t.first += 1000 * 1000 * 10;
    t.second += 1000 * 1000 * 10;
    v[i] = t;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) d[i][j] = -1;
  init();
  cout << solve(0, n - 1);
  return 0;
}
