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 L = 210;
const int M = 1000000007;
int x[L], y[L], f[L][L], n;
long long s[L][L];
bool in[L][L][L];
int get(int x) { return x % n == 0 ? n : x % n; }
long long cross(int a, int b, int c) {
  return (long long)(x[b] - x[a]) * (long long)(y[c] - y[a]) -
         (long long)(x[c] - x[a]) * (long long)(y[b] - y[a]);
}
long long absll(long long x) { return x > 0 ? x : -x; }
void init(void) {
  cin >> n;
  int i, j, k;
  for (i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++)
      for (k = i + 1; k < j; k++) s[i][j] += cross(k, k + 1, i);
  for (i = 1; i <= n; i++)
    for (j = 1; j < i; j++)
      for (k = get(i + 1); k != j; k = get(k + 1))
        s[i][j] += cross(get(k), get(k + 1), i);
}
void work(void) {
  int i, j, k;
  long long t;
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      for (k = j + 1; k <= n; k++) {
        t = cross(j, k, i);
        if (absll(t) + absll(s[i][j]) + absll(s[j][k]) + absll(s[k][i]) ==
                absll(s[1][n]) &&
            t)
          in[i][j][k] = 1;
      }
  for (i = 1; i <= n; i++) f[i][i + 1] = 1;
  for (i = n; i >= 1; i--)
    for (j = i + 1; j <= n; j++) {
      for (k = i + 1; k < j; k++)
        if (in[i][k][j])
          f[i][j] = ((long long)f[i][j] + (long long)f[i][k] * f[k][j]) % M;
    }
  printf("%d\n", f[1][n]);
}
int main(void) {
  init();
  work();
  return 0;
}
