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 + 5;
const int mod = 1000000007;
int N;
struct Point {
  int x, y;
  Point(int x = 0, int y = 0) : x(x), y(y) {}
  long long operator*(const Point &b) const {
    return (long long)x * b.y - (long long)y * b.x;
  }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  inline void scan() const { scanf("%d%d", &x, &y); }
} P[maxn];
long long dp[maxn][maxn];
long long solve(int l, int r) {
  if (~dp[l][r]) return dp[l][r];
  long long &ans = dp[l][r];
  ans = 0;
  if (r - l == 1) return ans = 1;
  for (int i = l + 1; i < r; i++)
    if ((P[l] - P[r]) * (P[i] - P[r]) > 0)
      (ans += (solve(l, i) * solve(i, r)) % mod) %= mod;
  return ans;
}
int main() {
  scanf("%d", &N);
  long long t = 0;
  for (int i = 1; i <= N; i++) P[i].scan();
  P[N + 1] = P[1];
  for (int i = 1; i <= N; i++) t += P[i] * P[i + 1];
  if (t < 0) reverse(P + 1, P + 1 + N);
  memset(dp, -1, sizeof dp);
  printf("%I64d", solve(1, N));
  return 0;
}
