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;
struct point {
  long long x, y;
  point() {}
  point(long long x, long long y) : x(x), y(y) {}
};
inline point toVec(point a, point b) { return point(b.x - a.x, b.y - a.y); }
inline long long cross(point a, point b) { return a.x * b.y - a.y * b.x; }
long long memo[202][202];
point pts[202];
long long dp(int l, int r) {
  if (r - l == 1) return 1;
  if (memo[l][r] == -1) {
    memo[l][r] = 0;
    for (int i = l + 1; i < r; i++) {
      if (cross(toVec(pts[l], pts[r]), toVec(pts[i], pts[r])) > 0)
        memo[l][r] = (memo[l][r] + dp(l, i) * dp(i, r)) % 1000000007;
    }
  }
  return memo[l][r];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(memo, -1, sizeof memo);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> pts[i].x >> pts[i].y;
  }
  long long area = 0;
  for (int i = 0; i < n; i++) {
    int j = (i + 1) % n;
    area += cross(pts[i], pts[j]);
  }
  if (area < 0) reverse(pts, pts + n);
  memset(memo, -1, sizeof memo);
  cout << dp(0, n - 1) << '\n';
}
