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 maxn = 1e6 + 50;
const long long inf = 0x3f3f3f3f3f3f3f3fLL;
struct node {
  long long x, y;
  node operator-(const node &a) const { return node{x - a.x, y - a.y}; }
  long long operator*(const node &a) const { return x * a.y - y * a.x; }
} my[550];
long long dp[250][250];
long long dfs(int l, int r) {
  if (l + 1 == r) return 1;
  if (~dp[l][r]) return dp[l][r];
  long long ans = 0;
  for (int i = l + 1; i < r; i++) {
    if ((my[l] - my[i]) * (my[i] - my[r]) <= 0) continue;
    ans += dfs(l, i) * dfs(i, r) % mod;
    ans %= mod;
  }
  return dp[l][r] = ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> my[i].x >> my[i].y;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += my[i] * my[(i + 1) % n];
  }
  if (ans < 0) {
    reverse(my, my + n);
  }
  memset(dp, -1, sizeof(dp));
  cout << dfs(0, n - 1) << endl;
  return 0;
}
