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 L {
  complex<long long> x, y;
  L();
  L(complex<long long> a, complex<long long> b) { x = a, y = b; };
};
inline long long cross(const complex<long long> &a,
                       const complex<long long> &b) {
  return a.real() * b.imag() - a.imag() * b.real();
}
inline long long dot(const complex<long long> &a, const complex<long long> &b) {
  return a.real() * b.real() + a.imag() * b.imag();
}
inline int ccw(const complex<long long> &a, const complex<long long> &b,
               const complex<long long> &c) {
  long long cr = cross(b - a, c - a);
  if (cr > 0) return 1;
  if (cr < -0) return -1;
  if (dot(b - a, c - a) <= 0) return 2;
  if (dot(a - b, c - b) <= 0) return -2;
  return 0;
}
bool insSS(const L &s, const L &t) {
  int a = ccw(s.x, s.y, t.x), b = ccw(s.x, s.y, t.y);
  int c = ccw(t.x, t.y, s.x), d = ccw(t.x, t.y, s.y);
  if (!a || !b || !c || !d) return true;
  if (a * b < 0 && c * d < 0) return true;
  return false;
}
int contains(const vector<complex<long long> > &pol, complex<long long> p) {
  bool in = false;
  for (int i = 0; i < pol.size(); i++) {
    complex<long long> a = pol[i] - p, b = pol[(i + 1) % pol.size()] - p;
    if (imag(a) > imag(b)) swap(a, b);
    if (imag(a) <= 0 && 0 < imag(b)) {
      if (cross(a, b) < 0) in = !in;
    }
  }
  return in;
}
const int MN = 220;
const int MD = 1e9 + 7;
int n;
complex<long long> p[MN];
vector<complex<long long> > pol;
long long dp[MN][MN][MN];
bool used[MN][MN][MN] = {};
vector<L> pl;
bool isok(int a, int b) {
  if (a == b) return false;
  if ((b + 1) % n == a || (a + 1) % n == b) return true;
  L l = L(p[a], p[b]);
  for (int i = 0; i < n; i++) {
    if (insSS(l, pl[i])) {
      return false;
    }
  }
  return contains(pol, (p[a] + p[b]) / 2LL);
}
bool iss[MN][MN] = {};
long long solve(int a, int b, int c) {
  if (c >= n || b >= c || a >= b) return 1;
  if (c - b == 1) return 1;
  if (used[a][b][c]) return dp[a][b][c];
  used[a][b][c] = true;
  long long r = 0;
  for (int i = b + 1; i <= c; i++) {
    if (iss[a][i] && iss[b][i]) {
      r += (solve(b, b + 1, i) * solve(a, i, c)) % MD;
      r %= MD;
    }
  }
  r %= MD;
  dp[a][b][c] = r;
  return r;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    double x, y;
    cin >> x >> y;
    p[i] = complex<long long>(x * 2, y * 2);
  }
  for (int i = 0; i < n; i++) {
    pl.push_back(L(p[i], p[(i + 1) % n]));
  }
  pol = vector<complex<long long> >(p, p + n);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      iss[i][j] = iss[j][i] = isok(i, j);
    }
  }
  cout << solve(0, 1, n - 1) << endl;
  return 0;
}
