Polycarp loves geometric progressions — he collects them. However, as such progressions occur very rarely, he also loves the sequences of numbers where it is enough to delete a single element to get a geometric progression.

In this task we shall define geometric progressions as finite sequences of numbers a1, a2, ..., ak, where ai = c·bi - 1 for some real numbers c and b. For example, the sequences [2, -4, 8], [0, 0, 0, 0], [199] are geometric progressions and [0, 1, 2, 3] is not.

Recently Polycarp has found a sequence and he can't classify it. Help him to do it. Determine whether it is a geometric progression. If it is not, check if it can become a geometric progression if an element is deleted from it.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of elements in the given sequence. The second line contains the given sequence. The numbers are space-separated. All the elements of the given sequence are integers and their absolute value does not exceed 104.

Output

Print 0, if the given sequence is a geometric progression. Otherwise, check if it is possible to make the sequence a geometric progression by deleting a single element. If it is possible, print 1. If it is impossible, print 2.

Examples

Input

4
3 6 12 24


Output

0


Input

4
-8 -16 24 -32


Output

1


Input

4
0 1 2 3


Output

2
#include <bits/stdc++.h>
using namespace std;
bool checkz(const vector<int>& v, int i) {
  return v[i] == 0 && v[i - 1] == 0 && v[i + 1] != 0;
}
int check(const vector<int>& v, bool b) {
  if (v.size() <= 2) {
    return !(v.size() == 1 || v[1] == 0 || v[0] != 0);
  }
  for (int i = 1; i < v.size() - 1; ++i) {
    if (v[i] * v[i] != v[i - 1] * v[i + 1] || checkz(v, i)) {
      if (b) {
        return 2;
      } else {
        b = true;
        int r = 2;
        for (int j = -1; j <= 1; ++j) {
          vector<int> v1 = v;
          v1.erase(v1.begin() + i + j);
          r = min(r, 1 + check(v1, true));
        }
        return r;
      }
    }
  }
  return 0;
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
  }
  cout << check(v, false) << endl;
}
