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;
const int mod = 1e9 + 7;
const double eps = 1e-8;
int n, f[N][N];
struct pt {
  int x, y;
  pt(int _x = 0, int _y = 0) {
    x = _x;
    y = _y;
  }
} p[N];
inline pt operator-(const pt &x, const pt &y) {
  return pt(x.x - y.x, x.y - y.y);
}
inline long long operator^(const pt &x, const pt &y) {
  return 1ll * x.x * y.y - 1ll * x.y * y.x;
}
inline void upd(int &x, int y) { x = (x + y >= mod) ? x + y - mod : x + y; }
inline int dp(int x, int y) {
  int &d = f[x][y];
  if (~d) return d;
  if (x + 1 >= y) return d = 1;
  d = 0;
  for (int i = x; i <= y; ++i)
    if (((p[i] - p[x]) ^ (p[y] - p[x])) < 0)
      upd(d, 1ll * dp(x, i) * dp(i, y) % mod);
  return d;
}
int main() {
  scanf("%d", &n);
  memset(f, -1, sizeof(f));
  for (int i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
  long long area = 0;
  for (int i = 2; i < n; ++i) area += (p[i] - p[1]) ^ (p[i + 1] - p[1]);
  if (area > 0) reverse(p + 2, p + n + 1);
  printf("%d\n", dp(1, n));
  return 0;
}
