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;
int a[111111];
int n;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  if (n == 1) {
    puts("0");
    return 0;
  }
  if (n == 2) {
    if (a[0] == 0 && a[1])
      puts("1");
    else
      puts("0");
    return 0;
  }
  int ans = 2;
  for (int k1 = 0; k1 < 3; ++k1) {
    for (int k2 = k1 + 1; k2 < 3; ++k2) {
      double q;
      if (a[k1] == 0 && a[k2] != 0) continue;
      if (a[k1] == 0)
        q = 0;
      else
        q = a[k2] / (a[k1] + 0.0);
      int p = k2;
      int cnt = 0;
      if (k1 != 0 || k2 != 1) cnt++;
      for (int i = k2 + 1; i < n; ++i) {
        if (fabs(a[p] * q - a[i]) < 1E-9) {
          p = i;
        } else {
          cnt++;
        }
      }
      ans = min(ans, cnt);
    }
  }
  cout << ans << endl;
}
