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 N = 210, P = 1e9 + 7;
int n;
int f[N][N];
struct Pt {
  long long x, y;
  long long val() { return x * x + y * y; }
  friend Pt operator-(const Pt &a, const Pt &b) {
    return {a.x - b.x, a.y - b.y};
  }
  friend bool operator==(const Pt &a, const Pt &b) {
    return a.x == b.x && a.y == b.y;
  }
  friend bool operator!=(const Pt &a, const Pt &b) { return !(a == b); }
} a[N];
inline int sgn(long long x) {
  if (x == 0) return 0;
  return x > 0 ? 1 : -1;
}
inline long long cross(const Pt &a, const Pt &b) {
  return a.x * b.y - a.y * b.x;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y);
  long long area = 0;
  for (int i = 1; i <= n; i++) area += cross(a[i], a[i % n + 1]);
  if (area < 0) reverse(a + 1, a + 1 + n);
  for (int i = 1; i < n; i++) f[i][i + 1] = 1;
  for (int k = 3; k <= n; k++)
    for (int i = 1; i + k - 1 <= n; i++) {
      for (int j = i + 1; j < i + k - 1; j++)
        if (cross(a[j] - a[i], a[i + k - 1] - a[i]) > 0)
          (f[i][i + k - 1] += 1LL * f[i][j] * f[j][i + k - 1] % P) %= P;
    }
  cout << f[1][n] << endl;
  return 0;
}
