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 maxn = 209;
const int MOD = 1e9 + 7;
struct point {
  long long x, y;
} p[maxn], t[maxn], o;
long long f[maxn][maxn], n;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long cross(point p, point a, point b) {
  return (a.x - p.x) * (b.y - p.y) - (a.y - p.y) * (b.x - p.x);
}
void init() {
  long long tmp = 0;
  o.x = 0;
  o.y = 0;
  p[n + 1] = p[1];
  for (int i = 1; i <= n; i++) tmp += cross(o, p[i], p[i + 1]);
  if (tmp < 0) {
    for (int i = 1; i <= n; i++) t[i] = p[n - i + 1];
    for (int i = 1; i <= n; i++) p[i] = t[i];
  }
}
int main() {
  int r;
  n = read();
  for (int i = 1; i <= n; i++) {
    p[i].x = read();
    p[i].y = read();
  }
  init();
  memset(f, 0, sizeof f);
  for (int i = 1; i <= n - 1; i++) f[i][i + 1] = 1;
  for (int len = 3; len <= n; len++)
    for (int l = 1; (r = l + len - 1) <= n; l++) {
      for (int k = l + 1; k <= r - 1; k++)
        if (cross(p[l], p[r], p[k]) < 0)
          f[l][r] = (f[l][r] + f[l][k] * f[k][r]) % MOD;
    }
  printf("%I64d\n", f[1][n] % MOD);
  return 0;
}
