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;
int f[410][410], n;
pair<double, double> a[410];
inline double cross(pair<double, double> x, pair<double, double> y,
                    pair<double, double> z) {
  return (y.first - x.first) * (z.second - x.second) -
         (y.second - x.second) * (z.first - x.first);
}
inline bool check() {
  double ans = 0;
  for (int i = 2; i <= n; i++) ans += cross(a[1], a[i - 1], a[i]);
  return ans > 0;
}
inline void revice() {
  for (int i = 1, j = n; i < j; i++, j--) swap(a[i], a[j]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].first, &a[i].second);
  if (check()) revice();
  for (int i = 1; i <= n; i++) a[i + n] = a[i];
  for (int i = 1; i + 1 <= n * 2; i++) f[i][i + 1] = 1;
  for (int l = 1; l <= n - 1; l++)
    for (int i = 1; i + l <= n * 2; i++)
      for (int j = i + 1; j < i + l; j++)
        (f[i][i + l] += f[i][j] * 1ll * f[j][i + l] % 1000000007 *
                        (cross(a[i], a[j], a[i + l]) < 0 ? 1 : 0)) %=
            1000000007;
  printf("%d\n", f[1][n]);
  return 0;
}
