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;
template <class T>
inline void read(T& x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
const int N = 200 + 5;
const int mo = 1000000007;
struct point {
  long long x, y;
  point() {}
  point(long long _x, long long _y) {
    x = _x;
    y = _y;
  }
} p[N];
point operator+(const point a, const point b) {
  return point(a.x + b.x, a.y + b.y);
}
point operator-(const point a, const point b) {
  return point(a.x - b.x, a.y - b.y);
}
long long operator*(const point a, const point b) {
  return a.x * b.y - a.y * b.x;
}
int n;
int a[N][N];
long long f[N][N];
int can(int i, int k, int j) { return (p[k] - p[i]) * (p[j] - p[i]) != 0; }
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(p[i].x), read(p[i].y);
  long long sum = 0;
  for (int i = 2; i <= n; i++) sum += p[i - 1] * p[i];
  sum += p[n] * p[1];
  if (sum < 0)
    for (int i = 1; i <= n / 2; i++) swap(p[i], p[n - i + 1]);
  for (int d = 1; d < n; d++) {
    for (int i = 1; i + d <= n; i++) {
      int j = i + d;
      if (d == 1) {
        f[i][j] = 1;
        continue;
      }
      for (int k = i + 1; k <= j - 1; k++) {
        if ((p[k] - p[i]) * (p[j] - p[i]) > 0) {
          (f[i][j] += f[i][k] * f[k][j]) %= mo;
        }
      }
    }
  }
  printf("%d\n", (f[1][n] + mo) % mo);
  return 0;
}
