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 MOD = 1000000007;
const int MAXN = 205;
int N;
complex<long long> P[MAXN];
long long dp[MAXN][MAXN];
long long cross_prod(complex<long long> p1, complex<long long> p2) {
  return imag(conj(p1) * p2);
}
long long getdp(int a, int b) {
  if (b - a == 1) return dp[a][b] = 1;
  if (dp[a][b] != -1) return dp[a][b];
  dp[a][b] = 0;
  for (int i = a + 1; i < b; i++)
    if (cross_prod(P[a] - P[i], P[b] - P[i]) < 0)
      dp[a][b] = (dp[a][b] + getdp(a, i) * getdp(i, b)) % MOD;
  return dp[a][b];
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    P[i] = complex<long long>(x, y);
  }
  long long area = 0;
  for (int i = 0; i < N; i++) area += cross_prod(P[i], P[(i + 1) % N]);
  if (area < 0) reverse(P, P + N);
  memset(dp, -1, sizeof(dp));
  cout << getdp(0, N - 1) << "\n";
  return 0;
}
