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;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  out << "(" << pair.first << ", " << pair.second << ")";
  return out;
}
long long skal(const complex<long long>& a, const complex<long long>& b) {
  return real(a) * real(b) + imag(a) * imag(b);
}
long long wek(const complex<long long>& a, const complex<long long>& b) {
  return imag(a) * real(b) - real(a) * imag(b);
}
bool lexi(const complex<long long>& a, const complex<long long>& b) {
  if (real(a) != real(b)) return real(a) < real(b);
  return imag(a) < imag(b);
}
const int N = 205;
const int MOD = 1e9 + 7;
int n;
complex<long long> pts[N];
bool e[N][N];
long long t[N][N];
int mod(int a) {
  if (a < n) return a;
  return a - n;
}
long long area(int a, int b) {
  long long res = wek(pts[b], pts[a]);
  for (int i = a; i != b; i = mod(i + 1)) {
    res += wek(pts[i], pts[mod(i + 1)]);
  }
  return abs(res);
}
int sgn(long long a) {
  if (a > 0) return 1;
  if (a < 0) return -1;
  return 0;
}
bool cross(complex<long long> a, complex<long long> b, complex<long long> c,
           complex<long long> d) {
  long long t1 = wek(c - a, b - a), t2 = wek(d - a, b - a);
  if (sgn(t1) * sgn(t2) > 0) return false;
  if (sgn(wek(a - c, d - c)) * sgn(wek(b - c, d - c)) > 0) return false;
  if (t1 == 0 && t2 == 0) {
    if (lexi(a, b)) swap(a, b);
    if (lexi(c, d)) swap(c, d);
    if (lexi(b, c) || lexi(d, a)) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    pts[i] = complex<long long>(x, y);
  }
  long long total = area(1, 0);
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j)
      if (i != j) {
        e[i][j] = true;
        for (int k = 0; k < (n); ++k) {
          int l = mod(k + 1);
          if (i == k || j == k || i == l || j == l) continue;
          if (cross(pts[i], pts[j], pts[k], pts[l])) e[i][j] = false;
        }
        long long a1 = area(i, j);
        long long a2 = area(j, i);
        if (a1 + a2 != total || a1 == 0 || a2 == 0) e[i][j] = false;
      }
  for (int i = 0; i < (n); ++i) {
    e[i][i] = true;
    e[i][mod(i + 1)] = true;
    e[mod(i + 1)][i] = true;
  }
  for (int i = 0; i < (n); ++i) {
    t[i][mod(i + 1)] = 1;
  }
  for (int l = (2); l <= (n - 1); ++l) {
    for (int i = 0; i < (n); ++i) {
      int j = mod(i + l);
      t[i][j] = 0;
      if (!e[i][j]) continue;
      for (int k = mod(i + 1); k != j; k = mod(k + 1)) {
        t[i][j] = (t[i][j] + t[i][k] * t[k][j]) % MOD;
      }
    }
  }
  cout << t[1][0] << endl;
  return 0;
}
