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;
struct node {
  int x, y;
} p[205];
int n, f[205][205], ok[205][205][205];
long long s[205][205];
long long cha(const node &a, const node &b, const node &c) {
  return 1ll * (c.x - a.x) * (b.y - a.y) - 1ll * (c.y - a.y) * (b.x - a.x);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &p[i].x, &p[i].y);
  for (int i = 1; i <= n; i++) {
    s[i][i] = 0;
    for (int j = i + 1; j <= n; j++) {
      s[i][j] = 0;
      for (int k = i + 1; k < j; k++) s[i][j] += cha(p[i], p[k], p[k + 1]);
      s[i][j] = (((s[i][j]) < 0) ? (-(s[i][j])) : (s[i][j]));
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) s[j][i] = s[1][n] - s[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int k = j + 1; k <= n; k++)
        if ((((cha(p[i], p[j], p[k])) < 0) ? (-(cha(p[i], p[j], p[k])))
                                           : (cha(p[i], p[j], p[k]))) +
                    s[i][j] + s[j][k] + s[k][i] ==
                s[1][n] &&
            cha(p[i], p[j], p[k]) != 0)
          ok[i][j][k] = 1;
  for (int i = 1; i < n; i++) f[i][i + 1] = 1;
  for (int i = 2; i < n; i++)
    for (int j = 1; j + i <= n; j++) {
      int k = j + i;
      f[j][k] = 0;
      for (int l = j + 1; l < k; l++)
        if (ok[j][l][k])
          f[j][k] =
              (f[j][k] + 1ll * f[j][l] * f[l][k] % 1000000007) % 1000000007;
    }
  printf("%d\n", f[1][n]);
  return 0;
}
