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;
int n;
complex<long long> p[300];
bool adj[300][300];
long long dp[300][300];
long long turn(complex<long long> a, complex<long long> b,
               complex<long long> c) {
  return real(a - b) * imag(c - b) - imag(a - b) * real(c - b);
}
long long dot(complex<long long> a, complex<long long> b) {
  return real(a) * real(b) + imag(a) * imag(b);
}
double angle(complex<long long> a, complex<long long> b, complex<long long> c) {
  long long s = turn(a, b, c);
  long long d = dot(a - b, c - b);
  return atan2(s, d);
}
bool cross(complex<long long> a, complex<long long> b, complex<long long> c,
           complex<long long> d) {
  long long t1 = turn(c, a, d);
  long long t2 = turn(c, b, d);
  long long t3 = turn(a, c, b);
  long long t4 = turn(a, d, b);
  if (t1 == 0 && t2 == 0 && t3 == 0 && t4 == 0) {
    return false;
  } else {
    if (a == c || a == d || b == c || b == d) return false;
    return ((t1 >= 0 && t2 <= 0) || (t1 <= 0 && t2 >= 0)) &&
           ((t3 >= 0 && t4 <= 0) || (t3 <= 0 && t4 >= 0));
  }
}
int main() {
  complex<long long> o = complex<long long>(0, 0);
  long long K = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    p[i] = complex<long long>(2 * x, 2 * y);
  }
  for (int i = 0; i < n; i++) {
    K += turn(p[i], o, p[(i + 1) % n]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      adj[i][j] = true;
      for (int k = 0; k < n; k++) {
        adj[i][j] &= !cross(p[i], p[j], p[k], p[(k + 1) % n]);
        if (turn(p[i], p[k], p[j]) == 0 && dot(p[i] - p[k], p[k] - p[j]) > 0) {
          adj[i][j] = false;
        }
      }
      if (j != (i + 1) % n && i != (j + 1) % n) {
        double K2 = 0;
        complex<long long> mid = p[i] + p[j];
        mid = complex<long long>(real(mid) / 2, imag(mid) / 2);
        for (int k = 0; k < n; k++) {
          K2 += angle(p[k], mid, p[(k + 1) % n]);
        }
        if (abs(K2) <= 1) adj[i][j] = false;
      }
      adj[j][i] = adj[i][j];
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j + i < n; j++) {
      if (!adj[j][j + i]) continue;
      if (i == 1) {
        dp[j][j + i] = 1;
        continue;
      }
      for (int k = j + 1; k < j + i; k++) {
        dp[j][j + i] += (dp[j][k] * dp[k][j + i]) % 1000000007;
      }
      dp[j][j + i] %= 1000000007;
    }
  }
  printf("%lld\n", dp[0][n - 1]);
}
