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 sqr(T x) {
  return x * x;
}
long long f[301][301];
const long long mod = 1e9 + 7;
pair<long long, long long> a[301];
int n;
pair<long long, long long> operator-(const pair<long long, long long> &x,
                                     const pair<long long, long long> &y) {
  return pair<long long, long long>(x.first - y.first, x.second - y.second);
}
long long operator*(const pair<long long, long long> &x,
                    const pair<long long, long long> &y) {
  return x.first * y.second - y.first * x.second;
}
inline long long xmult(pair<long long, long long> &x,
                       pair<long long, long long> &y) {
  return x.first * y.second - y.first * x.second;
}
long long dp(int l, int r) {
  if (f[l][r] >= 0) return f[l][r];
  if (l + 1 == r) return f[l][r] = 1;
  f[l][r] = 0;
  for (int i = l + 1; i <= r - 1; i++) {
    long long maki = (a[i] - a[l]) * (a[r] - a[i]);
    if (maki >= 0) continue;
    (f[l][r] += dp(l, i) * dp(i, r)) %= mod;
  }
  return f[l][r];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
  long long s = 0;
  for (int i = 0; i < n; i++) s += a[i] * a[(i + 1) % n];
  if (s > 0) reverse(a, a + n);
  memset(f, -1, sizeof(f));
  dp(0, n - 1);
  cout << f[0][n - 1] << endl;
}
