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, mod = 1000000007;
struct Point {
  long long x, y;
  void Read() { scanf("%lld%lld", &x, &y); }
  Point() {}
  Point(long long _x, long long _y) : x(_x), y(_y) {}
  friend Point operator-(const Point a, const Point b) {
    return Point(a.x - b.x, a.y - b.y);
  }
  friend long long operator*(const Point a, const Point b) {
    return a.x * b.y - a.y * b.x;
  }
} P[maxn];
int n;
int v[maxn][maxn];
long long f[maxn][maxn];
long long dp(int l, int r) {
  if (v[l][r]) return f[l][r];
  v[l][r] = 1;
  long long &res = f[l][r];
  if (l == r - 1) return res = 1;
  for (int k = l + 1; k < r; k++)
    if ((P[r] - P[l]) * (P[r] - P[k]) > 0) (res += dp(l, k) * dp(k, r)) %= mod;
  return res;
}
int main() {
  long long area = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) P[i].Read();
  P[n] = P[0];
  for (int i = 0; i < n; i++) area += P[i] * P[i + 1];
  if (area < 0) reverse(&P[0], &P[n]), P[n] = P[0];
  printf("%lld\n", dp(0, n - 1));
  return 0;
}
