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 long long mod = (int)(1e9) + 7;
const double eps = 1e-8;
int dcmp(double u) {
  if (u > eps) return 1;
  if (u < (-eps)) return -1;
  return 0;
}
struct Point {
  double x;
  double y;
  Point() {}
  Point(double u, double v) {
    x = u;
    y = v;
    return;
  }
  void input() {
    scanf("%lf%lf", &x, &y);
    return;
  }
  void output() { printf("%.9lf %.9lf ", x, y); }
  double len() { return sqrt(x * x + y * y); }
  void tolen(double u) {
    double l = len();
    x /= l;
    x *= u;
    y /= l;
    y *= u;
    return;
  }
  void rotate(double sx, double cx) {
    double x1 = x, y1 = y;
    x = x1 * cx - y1 * sx;
    y = x1 * sx + y1 * cx;
    return;
  }
};
Point er = Point(1e8, 1e9);
Point operator+(Point u, Point v) { return Point(u.x + v.x, u.y + v.y); }
Point operator-(Point u, Point v) { return Point(u.x - v.x, u.y - v.y); }
double det(Point u, Point v) { return u.x * v.y - u.y * v.x; }
long long f[201][201];
Point p[201];
int n;
long long solve(int x, int y) {
  if (y == x + 1 || y == x) return 1;
  if (f[x][y] != -1) return f[x][y];
  f[x][y] = 0;
  for (int i = x + 1; i < y; i++)
    if (dcmp(det(p[i] - p[x], p[i] - p[y])) < 0) {
      f[x][y] += solve(x, i) * solve(i, y);
      f[x][y] %= mod;
    }
  return f[x][y];
}
int main() {
  memset(f, -1, sizeof(f));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) p[i].input();
  double sum = 0;
  for (int i = 1; i < n; i++) sum += det(p[i + 1] - p[i], p[n] - p[i]);
  if (dcmp(sum) < 0) reverse(p + 1, p + n + 1);
  printf("%I64d\n", solve(1, n));
  return 0;
}
