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;
#pragma comment(linker, "/STACK:64000000")
template <class T>
inline T sqr(T x) {
  return x * x;
}
int n;
vector<int> a;
inline bool check(const vector<int> &a) {
  if ((int)((a).size()) == 1) return true;
  if (!a[0]) return count((a).begin(), (a).end(), 0) == (int)((a).size());
  long long a0 = a[0];
  long long a1 = a[1];
  for (int i = 1; i < (int)((a).size()); ++i) {
    if (a[i] * a0 != a[0] * a1) return false;
    a0 *= a[0];
    a1 *= a[1];
  }
  return true;
}
inline int check2(const vector<int> &a) {
  int n = (int)((a).size());
  if (n == 1) return true;
  if (!a[0]) {
    int cnt0 = count((a).begin(), (a).end(), 0);
    if (n - cnt0 > 1) return 2;
    if (n - cnt0 == 1) return 1;
    return 0;
  }
  long long a0 = a[0];
  long long a1 = a[1];
  bool first = true;
  for (int i = 1; i < n; ++i) {
    if (a[i] * a0 != a[0] * a1) {
      if (first) {
        first = false;
        continue;
      }
      return 2;
    }
    a0 *= a[0];
    a1 *= a[1];
  }
  return first ? 0 : 1;
}
int solve() {
  if (check(a)) return 0;
  if (check(vector<int>(a.begin() + 1, a.end()))) return 1;
  vector<int> tmp;
  tmp.push_back(a[0]);
  tmp.insert(tmp.end(), a.begin() + 2, a.end());
  if (check(tmp)) return 1;
  return check2(a);
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  printf("%d\n", solve());
  return 0;
}
