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;
long long n, dp[2005][2005];
struct node {
  long long x, y;
} a[2005];
long long operator*(node aa, node bb) { return aa.x * bb.y - aa.y * bb.x; }
node operator-(node aa, node bb) { return (node){aa.x - bb.x, aa.y - bb.y}; }
long long solve(long long l, long long r) {
  if (dp[l][r] != -1) return dp[l][r];
  long long &ans = dp[l][r];
  ans = 0;
  if (r - l == 1) return ans = 1;
  for (long long i = l + 1; i < r; i++)
    if ((a[l] - a[r]) * (a[i] - a[r]) > 0)
      ans = (ans + 1ll * solve(l, i) * solve(i, r) % 1000000007) % 1000000007;
  return ans;
}
signed main() {
  scanf("%lld", &n);
  memset(dp, -1, sizeof(dp));
  for (long long i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y);
  long long sum = 0;
  a[n + 1] = a[1];
  for (long long i = 1; i <= n; i++) sum += a[i] * a[i + 1];
  if (sum < 0) reverse(a + 1, a + n + 1);
  printf("%lld\n", solve(1, n));
  return 0;
}
