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 n;
const int mo = 1e9 + 7;
long long dp[205][205];
struct point {
  int x, y;
  void input() { scanf("%d%d", &x, &y); }
  point operator-(const point &a) {
    point ret;
    ret.x = x - a.x;
    ret.y = y - a.y;
    return ret;
  }
} p[205];
long long cross(point a, point b, point c) {
  return 1ll * (b.x - a.x) * (c.y - a.y) - 1ll * (b.y - a.y) * (c.x - a.x);
}
long long dfs(int l, int r) {
  if (l + 1 == r) {
    dp[l][r] = 1;
  }
  if (dp[l][r] != -1) return dp[l][r];
  dp[l][r] = 0;
  for (int i = l + 1; i < r; ++i) {
    if (cross(p[l], p[r], p[i]) >= 0) continue;
    dp[l][r] = (dp[l][r] + dfs(l, i) * dfs(i, r) % mo) % mo;
  }
  return dp[l][r];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    p[i].input();
  }
  double ans = 0;
  for (int i = 0; i < n - 1; ++i) {
    ans += cross(p[0], p[i], p[i + 1]);
  }
  if (ans < 0) {
    reverse(p, p + n);
  }
  memset(dp, -1, sizeof(dp));
  dfs(0, n - 1);
  printf("%d\n", (int)dp[0][n - 1]);
  return 0;
}
