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 long long N = 205, P = 1e9 + 7ll;
struct node {
  long long x, y;
  node operator-(const node &t) const {
    node c;
    c.x = x - t.x;
    c.y = y - t.y;
    return c;
  }
  long long operator*(const node &t) const { return x * t.y - y * t.x; }
} p[N];
long long n, dp[N][N];
long long search(long long x, long long y) {
  long long ans = 0;
  if (dp[x][y] != -1) return dp[x][y];
  if (x + 1 == y) return dp[x][y] = 1;
  for (int i = x + 1; i < y; i++)
    if ((p[x] - p[i]) * (p[y] - p[i]) > 0)
      ans = (ans + search(x, i) * search(i, y) % P) % P;
  return dp[x][y] = ans;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%dll", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  p[n + 1] = p[1];
  long long s = 0;
  for (int i = 1; i <= n; i++) s += p[i + 1] * p[i];
  if (s < 0)
    for (int i = 1, j = n; i < j; swap(p[i++], p[j--]))
      ;
  printf("%lld", search(1, n));
  return 0;
}
