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;
template <typename T>
void ri(T& x);
const int MAXN = 202, MOD = 1000000007;
int n, dir;
inline void inc(int& x) { x = (x + 1) % n; }
inline void dec(int& x) { x = (x - 1 + n) % n; }
inline int sign(long long x) { return (x > 0) ? 1 : ((x == 0) ? 0 : -1); }
struct V {
  long long x, y;
  long long operator*(const V& o) const { return x * o.y - y * o.x; }
  V operator-(const V& o) const { return V{x - o.x, y - o.y}; }
} p[MAXN];
long long f[MAXN][MAXN];
int main() {
  int kase = 1;
  while (kase--) {
    ri(n);
    for (int i = 0; i < n; ++i) {
      ri(p[i].x);
      ri(p[i].y);
    }
    long long area = 0;
    for (int i = 1, j = 2; j < n; ++i, ++j) {
      area += (p[i] - p[0]) * (p[j] - p[0]);
    }
    dir = sign(area);
    for (int i = 0; i + 1 < n; ++i) f[i][i + 1] = 1;
    for (int len = 2; len < n; ++len) {
      for (int i = 0; i + len < n; ++i) {
        const int j = i + len;
        f[i][j] = 0;
        for (int k = i + 1; k < j; ++k) {
          if (sign((p[k] - p[i]) * (p[j] - p[i])) == dir) {
            f[i][j] += f[i][k] * f[k][j] % MOD;
            f[i][j] %= MOD;
          }
        }
      }
    }
    cout << f[0][n - 1] << endl;
  }
  return 0;
}
void ri(int& x) {
  bool minus = false;
  char c;
  while ((c = getchar()) < '0' || '9' < c)
    if (c == '-') minus = true;
  x = c - '0';
  while ('0' <= (c = getchar()) && c <= '9') x = 10 * x + c - '0';
  if (minus) x = -x;
}
template <typename T>
void ri(T& x) {
  bool minus = false;
  char c;
  while ((c = getchar()) < '0' || '9' < c)
    if (c == '-') minus = true;
  x = c - '0';
  while ('0' <= (c = getchar()) && c <= '9') x = 10 * x + c - '0';
  if (minus) x = -x;
}
