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 oo = 2139063143;
const int N = 650;
const int P = 1000000007;
const double eps = 1e-7;
template <typename T>
inline void sc(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
template <typename T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void pr(T x) {
  print(x), putchar('\n');
}
template <typename T1, typename T2>
void chkmax(T1 &A, T2 B) {
  A = A >= B ? A : B;
}
template <typename T1, typename T2>
void chkmin(T1 &A, T2 B) {
  A = A <= B ? A : B;
}
int del(int x) { return x >= P ? x - P : x; }
void add(int &x, int y) { x = del(x + y); }
struct po {
  int x, y;
  po(int xx = 0, int yy = 0) { x = xx, y = yy; }
} a[N];
po operator+(po A, po B) { return po(A.x + B.x, A.y + B.y); }
po operator-(po A, po B) { return po(A.x - B.x, A.y - B.y); }
long long operator*(po A, po B) {
  return (long long)A.x * B.y - (long long)A.y * B.x;
}
int n, f[N][N];
int main() {
  sc(n);
  for (int i = 1; i <= n; i++) sc(a[i].x), sc(a[i].y);
  long long s = a[n] * a[1];
  for (int i = 1; i < n; i++) s += a[i] * a[i + 1];
  if (s < 0) reverse(a + 1, a + n + 1);
  for (int i = 1; i < n; i++) f[i][i + 1] = 1;
  for (int len = 3; len <= n; len++)
    for (int l = 1, r = len; r <= n; l++, r++)
      for (int x = l + 1; x < r; x++)
        if ((a[x] - a[l]) * (a[r] - a[l]) > 0)
          add(f[l][r], (long long)f[l][x] * f[x][r] % P);
  pr(f[1][n]);
  return 0;
}
