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>
const int N = 200 + 10, MOD = 1000000007;
int n;
std::pair<int, int> poly[N];
long long sum[N];
inline long long Cross(const std::pair<int, int> &a,
                       const std::pair<int, int> &b) {
  return (long long)a.first * b.second - (long long)a.second * b.first;
}
int mem[N][N];
inline long long Area(int l, int r) {
  return llabs(sum[r - 1] - sum[l - 1] + Cross(poly[r], poly[l]));
}
inline long long Area(const std::pair<int, int> &a,
                      const std::pair<int, int> &b,
                      const std::pair<int, int> &c) {
  return llabs(Cross(a, b) + Cross(b, c) + Cross(c, a));
}
int Solve(int l, int r) {
  int &res = mem[l][r];
  if (~res) return res;
  if (r - l < 2) return res = 1;
  res = 0;
  for (int i = l + 1; i < r; ++i) {
    long long temp = Area(poly[l], poly[i], poly[r]);
    if (temp && Area(l, i) + Area(i, r) + temp == Area(l, r))
      (res += (long long)Solve(l, i) * Solve(i, r) % MOD) %= MOD;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &poly[i].first, &poly[i].second);
  for (int i = 1; i <= n; ++i)
    sum[i] = sum[i - 1] + Cross(poly[i], poly[i % n + 1]);
  memset(mem, -1, sizeof mem);
  printf("%d\n", Solve(1, n));
  return 0;
}
