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;
template <class T1, class T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
  os << "{" << p.first << ", " << p.second << "}";
  return os;
}
int check(vector<long long>& v) {
  int n = v.size();
  long long c = v[0];
  double d;
  if (c == 0 && v[1] == 0)
    d = 0;
  else if (c == 0 && v[1] != 0)
    return 2;
  else
    d = v[1] / (c * 1.0l);
  double num = v[1];
  long long cnt = 0;
  for (int i = 2; i < n && cnt <= 2; ++i) {
    double old_num = num;
    num = num * d;
    if (fabs(v[i] - num) < 1E-10l) continue;
    cnt++;
    num = old_num;
  }
  return cnt;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n, 0);
  for (int i = 0; i < n; ++i) cin >> v[i];
  if (n <= 2) {
    if (n == 1)
      cout << 0 << endl;
    else {
      if (v[0] != 0) {
        cout << 0 << endl;
      } else {
        if (v[1] != 0)
          cout << 1 << endl;
        else
          cout << 0 << endl;
      }
    }
  } else {
    vector<long long> v1(n - 1, 0);
    for (int i = 1; i < n; ++i) v1[i - 1] = v[i];
    vector<long long> v2(n - 1, 0);
    v2[0] = v[0];
    for (int i = 2; i < n; ++i) v2[i - 1] = v[i];
    int ch0 = check(v);
    if (ch0 <= 1)
      cout << ch0 << endl;
    else {
      int ch1 = check(v1);
      int ch2 = check(v2);
      if (ch1 == 0 || ch2 == 0)
        cout << 1 << endl;
      else
        cout << 2 << endl;
    }
  }
  return 0;
}
