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 = 250, mod = 1000000007;
const double eps = 1e-7;
int dcmp(double x) {
  if (x > eps)
    return 1;
  else if (x < -eps)
    return -1;
  return 0;
}
struct Point {
  double x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  void scan() { scanf("%lf%lf", &x, &y); }
};
Point operator+(Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
Point operator*(Point A, double k) { return Point(A.x * k, A.y * k); }
Point operator/(Point A, double k) { return Point(A.x / k, A.y / k); }
double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; }
double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
int n, f[maxn][maxn] = {0};
Point O = Point(0, 0), p[maxn];
int Mul(long long a, long long b) { return (a *= b) >= mod ? a % mod : a; }
int Add(long long a, long long b) { return (a += b) >= mod ? a - mod : a; }
double Area() {
  int i;
  double s = 0;
  for (i = 0; i < n; i++) s += Cross(p[i + 1] - p[i], O - p[i]);
  return s;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 0; i < n; i++) p[i].scan(), f[i][i + 1] = 1;
  p[n] = *p;
  if (dcmp(Area()) < 0)
    for (i = 0; i < n / 2; i++) swap(p[i], p[n - i - 1]);
  for (i = 2; i < n; i++)
    for (j = 0; j < n; j++)
      for (k = j + 1; k < j + i; k++)
        if (dcmp(Cross(p[k] - p[j], p[j + i] - p[j])) == 1)
          f[j][j + i] = Add(f[j][j + i], Mul(f[j][k], f[k][j + i]));
  printf("%d\n", f[0][n - 1]);
  return 0;
}
