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 {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }
};
const int MAXN = 205, MOD = 1000 * 1000 * 1000 + 7;
Point p[MAXN];
int dp[MAXN][MAXN], n;
long long cp(const Point &a, const Point &b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
int getdp(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  dp[l][r] = 0;
  for (int i = l + 1; i + 1 <= r; i++)
    if (cp(p[r] - p[l], p[i] - p[l]) < 0)
      dp[l][r] = (dp[l][r] + (long long)getdp(l, i) * getdp(i, r)) % MOD;
  return dp[l][r];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].x >> p[i].y;
  p[n] = p[0];
  long long s = 0;
  for (int i = 0; i < n; i++) s += cp(p[i], p[i + 1]);
  if (s < 0) reverse(p, p + n + 1);
  for (int i = 0; i + 1 < n; i++) {
    dp[i][i + 1] = 1;
    for (int j = i + 2; j < n; j++) dp[i][j] = -1;
  }
  cout << getdp(0, n - 1) << '\n';
  return 0;
}
