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 modulo = 1e9 + 7;
const int max_n = 202;
struct point {
  long long x, y;
  point() {}
  point(long long _x, long long _y) : x(_x), y(_y) {}
};
point operator-(const point& p1, const point& p2) {
  return point(p1.x - p2.x, p1.y - p2.y);
}
point p[max_n];
int is_link[max_n][max_n];
int n;
int dp[max_n][max_n], f[max_n][max_n];
inline int mmult(int a, int b) { return (1LL * a * b) % modulo; }
inline int next_in_loop(int x) {
  ++x;
  if (x == n) x = 0;
  return x;
}
inline int msum(int a, int b) {
  int s = a + b;
  if (s >= modulo) s -= modulo;
  if (s < 0) s += modulo;
  return s;
}
long long vmult(const point& a, const point& b) {
  return a.x * b.y - a.y * b.x;
}
long long smult(const point& a, const point& b) {
  return a.x * b.x + a.y * b.y;
}
int point_in_segment(const point& p, const point& pl, const point& pr) {
  return vmult(pr - pl, p - pl) == 0 && smult(p - pl, pr - pl) >= 0 &&
         smult(p - pr, pl - pr) >= 0;
}
int find_dp(int l, int r) {
  if (f[l][r]) return dp[l][r];
  f[l][r] = 1;
  if (r == l + 1) {
    dp[l][r] = 1;
    return 1;
  }
  dp[l][r] = 0;
  for (int k = l + 1; k < r; ++k)
    if (vmult(p[k] - p[l], p[r] - p[l]) > 0)
      dp[l][r] = msum(dp[l][r], mmult(find_dp(l, k), find_dp(k, r)));
  return dp[l][r];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].x >> p[i].y;
  long long all_sq = 0;
  for (int i = 0; i < n; ++i) {
    int j = i - 1;
    if (j < 0) j += n;
    all_sq += vmult(p[j], p[i]);
  }
  if (all_sq < 0) {
    all_sq = -all_sq;
    reverse(p, p + n);
  }
  cout << find_dp(0, n - 1) << endl;
  return 0;
}
