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;
inline int rd() {
  int r;
  scanf("%d", &r);
  return r;
}
int vis[250][250], dp[250][250], x[250], y[250], n;
void inc(int &x, int y) { x = (x + y) % 1000000007; }
long long Cross(int a, int b, int c) {
  return 1LL * (x[a] - x[c]) * (y[b] - y[c]) -
         1LL * (x[b] - x[c]) * (y[a] - y[c]);
}
int F(int l, int r) {
  if (l + 1 == r) return 1;
  if (vis[l][r]) return dp[l][r];
  int ret = 0;
  vis[l][r] = 1;
  for (int i = l + 1; i < r; i++)
    if (Cross(l, i, r) > 0) inc(ret, 1LL * F(l, i) * F(i, r) % 1000000007);
  return dp[l][r] = ret;
}
void solve() {
  n = rd();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) vis[i][j] = dp[i][j] = 0;
  for (int i = 1; i <= n; i++) x[i] = rd(), y[i] = rd();
  long long sum = 0;
  for (int i = 2; i < n; i++) {
    sum += Cross(i, i + 1, 1);
  }
  if (sum < 0) {
    for (int i = 1; i <= n / 2; i++)
      swap(x[i], x[n - i + 1]), swap(y[i], y[n - i + 1]);
  }
  printf("%d\n", F(1, n));
  return;
}
int main() {
  solve();
  return 0;
}
