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 int N = 2e2 + 5;
const int mod = 1e9 + 7;
struct Point {
  long long int x, y;
  Point operator-(Point oth) { return (Point){x - oth.x, y - oth.y}; }
  long long int operator*(Point oth) { return x * oth.y - y * oth.x; }
} a[N];
vector<pair<Point, Point> > v;
int n, dp[N][N];
int mul(int x, int y) { return (long long int)x * y % mod; }
int add(int x, int y) { return (x + y) % mod; }
int f(int l, int r) {
  int &res = dp[l][r];
  if (res != -1) return res;
  if (l == r - 1) return res = 1;
  res = 0;
  for (int i = l + 1; i < r; i++)
    if ((a[i] - a[l]) * (a[r] - a[l]) > 0)
      res = add(res, mul(f(l, i), f(i, r)));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
  long long int area = 0;
  for (int i = 1; i < n; i++) area += a[i] * a[i + 1];
  area += a[n] * a[1];
  if (area < 0) reverse(a + 1, a + n + 1);
  memset(dp, -1, sizeof dp);
  cout << f(1, n);
}
