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 = 205, mod = 1e9 + 7;
struct pt {
  int x, y;
} p[N];
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline void inc(int& x, int y) { x = add(x, y); }
int f[N][N], n;
bool vis[N][N];
inline pt operator-(const pt& x, const pt& y) { return {x.x - y.x, x.y - y.y}; }
inline long long cross(const pt& x, const pt& y) {
  return 1ll * x.x * y.y - 1ll * x.y * y.x;
}
int dp(int l, int r) {
  if (vis[l][r]) return f[l][r];
  vis[l][r] = 1;
  if (l + 1 >= r) return f[l][r] = 1;
  int& d = f[l][r];
  d = 0;
  for (int i = l + 1; i < r; ++i)
    if (cross(p[l] - p[r], p[i] - p[r]) > 0)
      inc(d, 1ll * dp(l, i) * dp(i, r) % mod);
  return d;
}
int main() {
  scanf("%d", &n);
  long long s = 0;
  for (int i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
  for (int i = 2; i < n; ++i) s += cross(p[i] - p[1], p[i + 1] - p[1]);
  if (s < 0) reverse(p + 2, p + n + 1);
  printf("%d\n", dp(1, n));
  return 0;
}
