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 N = 205;
const long long M = 1e9 + 7;
long long dp[N][N];
struct P {
  long long x, y;
  P() {}
  P(long long x, long long y) : x(x), y(y) {}
} p[N];
inline long long cal(P a, P b) { return a.x * b.y - a.y * b.x; }
inline P Pc(P a, P b) { return P(a.x - b.x, a.y - b.y); }
inline void re(long long n) {
  long long i = 0, j = n - 1;
  while (i < j) {
    P t = p[i];
    p[i] = p[j];
    p[j] = t;
    i++;
    j--;
  }
  return;
}
int main() {
  long long n, sum = 0;
  scanf("%lld", &n);
  for (long long i = 0; i < n; ++i) scanf("%lld%lld", &p[i].x, &p[i].y);
  for (long long i = 1; i < n - 1; ++i)
    sum += cal(Pc(p[i], p[0]), Pc(p[i + 1], p[0]));
  if (sum < 0) re(n);
  for (long long i = 0; i < n; ++i) dp[i][(i + 1) % n] = 1;
  for (long long l = 2; l <= n; ++l) {
    for (long long i = 0; i < n; ++i) {
      long long j = (i + l) % n;
      for (long long k = (i + 1) % n; k != j; (++k) %= n)
        if (cal(Pc(p[k], p[i]), Pc(p[j], p[i])) > 0)
          (dp[i][j] += dp[i][k] * dp[k][j]) %= M;
    }
  }
  printf("%lld\n", dp[0][n - 1]);
  return 0;
}
