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;
const int maxn = 100000 + 10;
int seq[maxn], bk[maxn], n;
bool check() {
  if (n - 1 <= 2) return true;
  for (int i = 1; i < n - 2; i++)
    if (bk[i - 1] * bk[i + 1] != bk[i] * bk[i] || bk[i] == 0 && bk[i + 1] != 0)
      return false;
  return true;
}
int solve() {
  if (n < 2) return 0;
  if (n == 2) {
    if (seq[0] == 0 && seq[1] != 0) return 1;
    return 0;
  }
  int p;
  for (p = 1; p < n - 1; p++)
    if (seq[p - 1] * seq[p + 1] != seq[p] * seq[p] ||
        seq[p] == 0 && seq[p + 1] != 0)
      break;
  if (p == n - 1) return 0;
  copy(seq, seq + p - 1, bk);
  copy(seq + p, seq + n, bk + p - 1);
  if (check()) return 1;
  bk[p - 1] = seq[p - 1];
  if (check()) return 1;
  bk[p] = seq[p];
  if (check()) return 1;
  return 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &seq[i]);
  cout << solve() << endl;
  return 0;
}
