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 mod = 1000000007;
struct point {
  long long x, y;
  point() {}
  point(long long _x, long long _y) {
    x = _x;
    y = _y;
  }
  long long operator^(const point &b) const { return x * b.y - y * b.x; }
  point operator-(const point &b) const { return point(x - b.x, y - b.y); }
} p[210];
long long dp[210][210];
long long dfs(int l, int r) {
  if (dp[l][r] >= 0) return dp[l][r];
  if (l == r - 1) return 1;
  long long res = 0;
  for (int i = l + 1; i < r; i++) {
    if (((p[l] - p[r]) ^ (p[i] - p[l])) > 0) {
      res = (res + dfs(l, i) * dfs(i, r) % mod) % mod;
    }
  }
  return dp[l][r] = res;
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%I64d%I64d", &p[i].x, &p[i].y);
  long long s = 0;
  for (int i = 0; i < n; i++) s += p[i] ^ p[(i + 1) % n];
  if (s < 0) reverse(p, p + n);
  memset(dp, -1, sizeof(dp));
  long long ans = dfs(0, n - 1);
  printf("%I64d\n", ans);
}
