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>
#pragma warning(disable : 4996)
using namespace std;
long long n;
pair<long long, long long> a[300];
long long d[300][300];
long long ccw(pair<long long, long long> a, pair<long long, long long> b,
              pair<long long, long long> c) {
  b.first -= a.first, b.second -= a.second;
  c.first -= a.first, c.second -= a.second;
  return b.first * c.second - b.second * c.first;
}
int main() {
  long long i, j, k;
  scanf("%I64d", &n);
  for (i = 0; i < n; i++) scanf("%I64d%I64d", &a[i].first, &a[i].second);
  a[n].first = a[0].first, a[n].second = a[0].second;
  long long ori = 0;
  for (i = 0; i < n; i++) {
    ori += (a[i + 1].first - a[i].first) * (a[i + 1].second + a[i].second);
  }
  if (ori > 0) {
    reverse(a, a + n + 1);
  }
  memset(d, -1, sizeof(d));
  for (i = 0; i < n - 1; i++) d[i][i + 1] = 1;
  for (k = 2; k < n; k++) {
    for (i = 0; i < n; i++) {
      if (i + k >= n) continue;
      long long sum = 0;
      for (j = i + 1; j < i + k; j++) {
        if (ccw(a[i], a[i + k], a[j]) >= 0) continue;
        sum += d[i][j] * d[j][i + k];
        sum %= 1000000007;
      }
      d[i][i + k] = sum;
    }
  }
  printf("%I64d", d[0][n - 1]);
  return 0;
}
