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;
struct Pt {
  long long x, y;
};
long long area(Pt a, Pt b, Pt c) {
  long long s = (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
  return s;
}
bool ccw(Pt a, Pt b, Pt c) { return area(a, b, c) > 0; }
int n;
inline int next(int x) { return x + 1 == n ? 0 : x + 1; }
int main() {
  cin >> n;
  vector<Pt> pts(n);
  for (int i = 0; i < n; ++i) cin >> pts[i].x >> pts[i].y;
  long long s_area = 0;
  for (int i = 0; i < n; ++i) s_area += area({0, 0}, pts[i], pts[next(i)]);
  if (s_area < 0) reverse(pts.begin(), pts.end());
  const int MAXN = 210;
  const long long MOD = 1e9 + 7;
  static long long dp[MAXN][MAXN];
  for (int i = 0; i < n; ++i) dp[i][next(i)] = 1;
  for (int len = 2; len < n; ++len) {
    for (int i = 0; i < n; ++i) {
      int j = (i + len) % n;
      dp[i][j] = 0;
      for (int s = 1; s < len; ++s) {
        int k = (i + s) % n;
        if (ccw(pts[i], pts[k], pts[j]) > 0) {
          dp[i][j] += dp[i][k] * dp[k][j];
          dp[i][j] %= MOD;
        }
      }
    }
  }
  int ans = dp[0][n - 1] % MOD;
  cout << ans << endl;
  return 0;
}
