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;
int n, v[205][205];
long long dp[205][205];
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  PT(const PT &p) : x(p.x), y(p.y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(double c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
};
double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
double ComputeSignedArea(const vector<PT> &p) {
  double area = 0;
  for (int i = 0; i < p.size(); i++) {
    int j = (i + 1) % p.size();
    area += p[i].x * p[j].y - p[j].x * p[i].y;
  }
  return area / 2.0;
}
vector<PT> poly;
double tArea;
long long solve(int f, int t) {
  if (f + 1 == t) return 1LL;
  if (v[f][t]) return dp[f][t];
  v[f][t] = 1;
  long long r = 0LL, aux;
  for (int i = f + 1; i < t; i++) {
    if (cross(poly[f] - poly[i], poly[t] - poly[i]) < 0) {
      aux = (solve(f, i) * solve(i, t)) % 1000000007;
      r = (r + aux) % 1000000007;
    }
  }
  return dp[f][t] = r;
}
int main() {
  PT point;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> point.x >> point.y;
    poly.push_back(point);
  }
  tArea = ComputeSignedArea(poly);
  if (tArea < 0) reverse(poly.begin(), poly.end());
  cout << solve(0, n - 1) << endl;
  return 0;
}
