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 = 200 + 8;
const long long modd = 1e9 + 7;
struct Point {
  long long x, y;
  long long operator^(const Point& b) { return x * b.y - b.x * y; }
} p[N];
int dp[N][N];
long long cross(Point a, Point b, Point c) {
  return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
long long dfs(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l + 1 == r) return dp[l][r] = 1;
  long long ans = 0;
  for (int k = l + 1; k < r; ++k) {
    if (cross(p[l], p[k], p[r]) > 0)
      ans = (ans + dfs(l, k) * dfs(k, r) % modd) % modd;
  }
  return dp[l][r] = ans;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld %lld", &p[i].x, &p[i].y);
  p[n + 1] = p[1];
  p[n + 2] = p[2];
  long long tem = 0;
  for (int i = 1; i <= n; ++i) tem += (p[i] ^ p[i + 1]);
  if (tem < 0) reverse(p + 1, p + 1 + n);
  printf("%lld", dfs(1, n));
  return 0;
}
