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 = 1e9 + 7;
const int N = 205;
struct Point {
  long long x, y;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  void read() { scanf("%lld%lld", &x, &y); }
} p[N];
Point operator-(const Point &a, const Point &b) {
  return Point(a.x - b.x, a.y - b.y);
}
long long operator*(const Point &a, const Point &b) {
  return a.x * b.y - a.y * b.x;
}
int n;
long long dp[N][N];
void pre_init() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].read();
  memset(dp, -1, sizeof(dp));
  long long tmp = 0;
  p[n] = p[0];
  for (int i = 0; i < n; i++) {
    tmp += p[i] * p[i + 1];
  }
  if (tmp < 0) reverse(p, p + n);
}
long long dfs(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  long long &ans = dp[l][r];
  ans = 0;
  if (r - l == 1) return ans = 1;
  for (int i = l + 1; i < r; i++) {
    if ((p[l] - p[r]) * (p[i] - p[r]) > 0) {
      ans = (ans + dfs(l, i) * dfs(i, r) % mod) % mod;
    }
  }
  return ans;
}
int main() {
  pre_init();
  cout << dfs(0, n - 1) << endl;
}
