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;
template <class T>
T read1() {
  T t = 0;
  bool vis = 0;
  char k;
  do (k = getchar()) == '-' && (vis = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return vis ? -t : t;
}
template <class T>
void print(T arg1) {
  cout << arg1;
}
template <class T>
void get(T arg1) {
  cout << arg1;
}
template <typename T, typename... Ts>
void print(T arg1, Ts... arg_left) {
  cout << arg1 << " ";
  print(arg_left...);
}
template <typename T, typename... Ts>
void get(T arg1, Ts... arg_left) {
  cout << arg1 << " ";
  print(arg_left...);
}
pair<int, int> a[205];
int s, dp[205][205];
pair<int, int> operator-(const pair<int, int> &x, const pair<int, int> &y) {
  return pair<int, int>(x.first - y.first, x.second - y.second);
}
long long operator*(const pair<int, int> &x, const pair<int, int> &y) {
  return 1ll * x.first * y.second - 1ll * x.second * y.first;
}
int main() {
  s = read1<int>();
  long long t = 0;
  for (int i = 1; i <= s; ++i) {
    a[i] = make_pair(read1<int>(), read1<int>());
    if (i != 1) t += a[i - 1] * a[i];
  }
  t += a[s] * a[1];
  for (int i = 1; i < s; ++i) dp[i][i + 1] = 1;
  if (t > 0) reverse(a + 1, a + s + 1);
  for (int i = s; --i;)
    for (int j = i + 2; j <= s; ++j)
      for (int k = i + 1; k < j; ++k)
        if ((a[j] - a[i]) * (a[k] - a[i]) > 0)
          dp[i][j] = (dp[i][j] + 1ll * dp[i][k] * dp[k][j]) % 1000000007;
  printf("%d", dp[1][s]);
  return 0;
}
