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 = 203;
const int MOD = (int)1e9 + 7;
struct Point {
  int x, y;
  Point(int _x = 0, int _y = 0) : x(_x), y(_y) {}
  long long operator*(const Point &o) const {
    return (long long)x * o.y - (long long)y * o.x;
  }
  Point operator-(const Point &o) const { return Point(x - o.x, y - o.y); }
};
int n;
Point A[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &A[i].x, &A[i].y);
  long long area = 0;
  A[n] = A[0];
  for (int i = 0; i < n; i++) area += A[i] * A[i + 1];
  if (area < 0) reverse(A, A + n);
  static long long dp[MAXN][MAXN];
  for (int len = 1; len < n; len++)
    for (int i = 0; i < n; i++) {
      int j = i + len;
      if (j >= n) break;
      if (len == 1) {
        dp[i][j] = 1;
        continue;
      }
      for (int k = i + 1; k < j; k++) {
        if ((A[i] - A[j]) * (A[k] - A[j]) > 0) {
          dp[i][j] += dp[i][k] * dp[k][j];
          dp[i][j] %= MOD;
        }
      }
    }
  cout << dp[0][n - 1] << endl;
  return 0;
}
