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 = 210;
const long long mod = 1000000007;
long long read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
struct point {
  long long x, y;
};
int n;
point p[maxn];
long long f[maxn][maxn];
long long operator^(point p1, point p2) { return p1.x * p2.y - p2.x * p1.y; }
point operator-(point p1, point p2) {
  return (point){p1.x - p2.x, p1.y - p2.y};
}
void init() {
  int i, j;
  long long x = 0;
  p[n + 1] = p[1];
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) f[i][j] = -1;
  for (i = 1; i <= n; i++) f[i][i + 1] = 1;
  for (i = 1; i <= n; i++) x += p[i] ^ p[i + 1];
  if (x < 0) reverse(p + 1, p + n + 2);
}
long long dfs(int l, int r) {
  if (f[l][r] != -1) return f[l][r];
  int i;
  f[l][r] = 0;
  for (i = l + 1; i < r; i++)
    if (((p[l] - p[r]) ^ (p[i] - p[r])) > 0)
      f[l][r] = (f[l][r] + dfs(l, i) * dfs(i, r) % mod) % mod;
  return f[l][r];
}
int main() {
  int i, j, k;
  cin >> n;
  for (i = 1; i <= n; i++) p[i].x = read(), p[i].y = read();
  init();
  cout << dfs(1, n) << endl;
  return 0;
}
