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 Size = 210;
const long long Mod = 1000000007;
struct Point {
  long long X, Y;
  long long operator*(const Point& A) const { return X * A.Y - A.X * Y; }
  Point operator-(const Point& A) const {
    Point P;
    P.X = X - A.X, P.Y = Y - A.Y;
    return P;
  }
};
int N;
Point P[Size];
long long F[Size][Size];
void Init() {
  scanf("%d", &N);
  memset(F, -1, sizeof(F));
  for (int i = 0; i < N; ++i) scanf("%I64d%I64d", &P[i].X, &P[i].Y);
  P[N] = P[0];
  long long Tmp = 0;
  for (int i = 0; i < N; ++i) Tmp += P[i] * P[i + 1];
  if (Tmp < 0) reverse(P, P + N);
}
long long DP(int L, int R) {
  if (F[L][R] != -1) return F[L][R];
  long long& Res = F[L][R];
  Res = 0;
  if (R - L == 1) return Res = 1;
  for (int i = L + 1; i < R; ++i) {
    if ((P[L] - P[R]) * (P[i] - P[R]) > 0)
      Res = (Res + DP(L, i) * DP(i, R)) % Mod;
  }
  return Res;
}
int main() {
  Init();
  printf("%I64d\n", DP(0, N - 1));
}
