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 mod = 1e9 + 7;
const int N = 205;
inline void cm(long long& x) {
  if (x >= mod) x %= mod;
}
inline int SIGNAL(long long x) {
  if (x < 0)
    return -1;
  else if (x > 0)
    return 1;
  else
    return 0;
}
struct pnt {
  long long x, y;
  pnt() {}
  pnt(long long a, long long b) { x = a, y = b; }
  void scan() { scanf("%lld%lld", &x, &y); }
  long long operator*(const pnt& a) { return x * a.y - a.x * y; }
  pnt operator-(const pnt& a) { return pnt(x - a.x, y - a.y); }
} pt;
vector<pnt> poly;
int flag, n;
void getflag() {
  long long sum = poly[n - 1] * poly[0];
  for (int i = 0; i < n - 1; ++i) sum += poly[i] * poly[i + 1];
  flag = SIGNAL(sum);
}
long long dp[N][N];
long long solve(int l, int r) {
  if (l + 1 == r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  long long res = 0;
  for (int i = l + 1; i < r; ++i) {
    if (SIGNAL((poly[r] - poly[l]) * (poly[r] - poly[i])) == flag) {
      res += solve(l, i) * solve(i, r);
      cm(res);
    }
  }
  return dp[l][r] = res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    pt.scan();
    poly.push_back(pt);
  }
  getflag();
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", solve(0, n - 1));
}
