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;
static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const double PI2 = 8.0 * atan(1.0);
int check(const vector<long long>& as) {
  if (as.size() <= 1) {
    return 0;
  }
  double c = as[0];
  if (c == 0) {
    return as.size() - count((as).begin(), (as).end(), 0);
  }
  double b = as[1] / c;
  int skipped = 0;
  double expected = c * b * b;
  for (int index = 2; index < as.size(); ++index) {
    if (abs(expected - as[index]) < EPS) {
      expected *= b;
    } else {
      ++skipped;
    }
  }
  return skipped;
}
int main() {
  int N;
  cin >> N;
  vector<long long> as;
  for (int n = 0; n < (int)N; ++n) {
    long long a;
    cin >> a;
    as.push_back(a);
  }
  int answer = 2;
  answer = min(answer, check(as));
  {
    vector<long long> bs = as;
    bs.erase(bs.begin(), bs.begin() + 1);
    answer = min(answer, check(bs) + 1);
  }
  if (N > 1) {
    vector<long long> cs = as;
    cs.erase(cs.begin() + 1, cs.begin() + 2);
    answer = min(answer, check(cs) + 1);
  }
  cout << answer << endl;
}
