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 = 205;
const long long mod = 1e9 + 7;
struct point {
  long long x, y;
  long long operator*(const point a) const { return x * a.y - y * a.x; }
  point operator-(const point a) const {
    point ans;
    ans.x = x - a.x;
    ans.y = y - a.y;
    return ans;
  }
} p[maxn + 1];
long long n, a = 0, d[maxn + 1][maxn + 1];
void reloc() {
  freopen("/Users/admin/Desktop/cards.in", "r", stdin);
  freopen("/Users/admin/Desktop/cards.out", "w", stdout);
  return;
}
long long dp(long long l, long long r) {
  if (d[l][r] != -1) return d[l][r];
  long long &x = d[l][r];
  x = 0;
  if (r - l == 1) return x = 1;
  for (long long i = l + 1; i < r; i++) {
    if ((p[l] - p[r]) * (p[i] - p[r]) > 0) {
      x = (x + dp(l, i) * dp(i, r)) % mod;
    }
  }
  return x;
}
int main() {
  memset(d, -1, sizeof(d));
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld %lld", &p[i].x, &p[i].y);
  }
  p[n + 1] = p[1];
  for (int i = 1; i <= n; i++) {
    a += p[i] * p[i + 1];
  }
  if (a < 0) reverse(p + 1, p + 1 + n), p[n + 1] = p[1];
  printf("%lld\n", dp(1, n));
  return 0;
}
