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;
struct P {
  long long x, y;
  P(long long _ = 0, long long $ = 0) { x = _, y = $; }
} p[205];
P operator-(P _, P $) { return P(_.x - $.x, _.y - $.y); }
long long operator*(P _, P $) { return _.x * $.y - _.y * $.x; }
int dp[205][205], n;
int dfs(int l, int r) {
  if (l + 1 == r) return 1;
  if (~dp[l][r]) return dp[l][r];
  for (int k = l + 1; k < r; k++)
    if ((p[l] - p[r]) * (p[k] - p[r]) > 0)
      dp[l][r] = (dp[l][r] + 1ll * dfs(l, k) * dfs(k, r)) % 1000000007;
  return ++dp[l][r];
}
int main() {
  scanf("%d", &n);
  long long s = 0;
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  for (int i = 2; i < n; i++) s += (p[i] - p[1]) * (p[i + 1] - p[1]);
  if (s < 0) reverse(p + 2, p + 1 + n);
  printf("%d", dfs(1, n));
}
