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 point {
  long long x, y;
  point() {}
  point(long long _x, long long _y) { x = _x, y = _y; }
};
long long crossProduct(point A, point B) { return (A.x * B.y) - (B.x * A.y); }
bool isLeftTurn(point A, point B, point C) {
  long long get =
      (crossProduct(A, B) + crossProduct(B, C) + crossProduct(C, A));
  return (get < 0ll);
}
bool clockWise(vector<point> &V) {
  long long val = 0ll;
  int sz = V.size();
  for (int i = 0; i < sz; i++) {
    val += crossProduct(V[i], V[(i + 1) % sz]);
  }
  return (val > 0ll);
};
vector<point> V;
long long dp[222][222];
long long solve(int l, int r) {
  if (r == l + 1) return 1ll;
  if (l == r) return 0ll;
  long long &ret = dp[l][r];
  if (ret != -1) return ret;
  ret = 0ll;
  for (int i = l + 1; i < r; i++) {
    if (isLeftTurn(V[l], V[i], V[r]))
      ret = (ret + ((solve(l, i) % 1000000007) * (solve(i, r) % 1000000007)) %
                       1000000007) %
            1000000007;
  }
  return ret;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    V.push_back(point(x, y));
  }
  if (clockWise(V)) reverse(V.begin(), V.end());
  memset(dp, -1, sizeof dp);
  long long get = solve(0, n - 1);
  cout << get << endl;
  return 0;
}
