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 MAXN = 205, MOD = 1e9 + 7;
struct point {
  long long x, y;
  point() {}
  point(long long _x, long long _y) : x(_x), y(_y) {}
  friend point operator-(const point a, const point b) {
    return point(a.x - b.x, a.y - b.y);
  }
  friend long long operator*(const point a, const point b) {
    return a.x * b.y - a.y * b.x;
  }
} pts[MAXN];
int n;
bool vis[MAXN][MAXN];
long long f[MAXN][MAXN];
long long dp(int l, int r) {
  if (vis[l][r]) return f[l][r];
  vis[l][r] = true;
  long long& res = f[l][r];
  if (l == r - 1) return res = 1;
  for (int k = l + 1; k < r; ++k)
    if ((pts[r] - pts[l]) * (pts[r] - pts[k]) > 0) {
      res += dp(l, k) * dp(k, r);
      res %= MOD;
    }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> pts[i].x >> pts[i].y;
  pts[n] = pts[0];
  long long area = 0;
  for (int i = 0; i < n; ++i) area += pts[i] * pts[i + 1];
  if (area < 0) reverse(pts, pts + n), pts[n] = pts[0];
  cout << dp(0, n - 1) << endl;
}
