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 = 200 + 10;
const int mod = 1000000007;
struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
  Point operator-(const Point& A) { return Point(x - A.x, y - A.y); }
};
Point v[maxn];
long long Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
bool Isclockwise(int n) {
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += Cross(v[i], v[i + 1]);
  return ans < 0;
}
int n;
long long d[maxn][maxn];
long long dp(int l, int r) {
  if (d[l][r] != -1) return d[l][r];
  long long& ans = d[l][r];
  ans = 0;
  if (r == l + 1) return ans = 1;
  for (int i = l + 1; i < r; i++) {
    if (Cross(v[r] - v[l], v[i] - v[l]) > 0)
      ans = (ans + dp(l, i) * dp(i, r)) % mod;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d%I64d", &v[i].x, &v[i].y);
  v[n + 1] = v[1];
  if (!Isclockwise(n)) reverse(v + 1, v + n + 1);
  memset(d, -1, sizeof(d));
  long long ans = dp(1, n);
  printf("%I64d\n", ans);
  return 0;
}
