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;
inline int min(int _a, int _b) { return _a < _b ? _a : _b; }
inline int max(int _a, int _b) { return _a > _b ? _a : _b; }
template <class T>
inline void rd(T &_a) {
  int _f = 0, _ch = getchar();
  _a = 0;
  while (_ch < '0' || _ch > '9') {
    if (_ch == '-') _f = 1;
    _ch = getchar();
  }
  while (_ch >= '0' && _ch <= '9')
    _a = (_a << 1) + (_a << 3) + _ch - '0', _ch = getchar();
  if (_f) _a = -_a;
}
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int N = 205, mod = 1e9 + 7;
struct Pt {
  long long x, y;
} p[N];
long long dp[N][N];
long long cmul(Pt p0, Pt p1, Pt p2) {
  return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);
}
long long sl(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l + 1 == r) return dp[l][r] = 1;
  long long &res = dp[l][r];
  res = 0;
  for (int i = l + 1; i < r; i++)
    if (cmul(p[l], p[i], p[r]) > 0) (res += sl(l, i) * sl(i, r)) %= mod;
  return res;
}
int main() {
  memset(dp, -1, sizeof dp);
  int n;
  long long s = 0;
  rd(n);
  for (int i = 0; i < n; i++) rd(p[i].x), rd(p[i].y);
  for (int i = 0; i < n; i++) s += cmul((Pt){0, 0}, p[i], p[(i + 1) % n]);
  if (s < 0) reverse(p, p + n);
  printf("%I64d", sl(0, n - 1));
  return 0;
}
