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;
using ll = long long;
int const N = 222;
int const M = 1e9 + 7;
int n;
ll f[N][N];
struct point {
  ll x, y;
  point(ll _x = 0, ll _y = 0) : x(_x), y(_y) {}
  ll operator*(const point &rhs) const { return x * rhs.y - y * rhs.x; }
  point operator-(const point &rhs) const {
    return point(x - rhs.x, y - rhs.y);
  }
} p[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%I64d%I64d", &p[i].x, &p[i].y);
  ll area = 0;
  for (int i = 1; i + 1 < n; ++i) area += (p[i] - p[0]) * (p[i + 1] - p[0]);
  if (area < 0) reverse(p, p + n);
  for (int i = 0; i < n; ++i) f[i][i] = f[i][(i + 1) % n] = 1;
  for (int step = 2; step < n; ++step)
    for (int i = 0; i < n; ++i) {
      int j = (i + step) % n;
      for (int t = 1; t <= step - 1; ++t) {
        int k = (i + t) % n;
        if ((p[k] - p[i]) * (p[j] - p[i]) > 0) {
          f[i][j] = (f[i][j] + f[i][k] * f[k][j]) % M;
        }
      }
    }
  printf("%I64d\n", f[0][n - 1]);
  return 0;
}
