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 n, i, j, a[111111], m = 2, t;
int main() {
  cin >> n;
  a[n] = 111111111;
  if (n == 1) {
    cout << "0\n";
    return 0;
  }
  for (i = 0; i < n; i++) cin >> a[i];
  if (n == 2) {
    if (!a[0] && a[1])
      cout << "1\n";
    else
      cout << "0\n";
    return 0;
  }
  for (i = 1; i < n; i++) t = t + (bool)a[i];
  if (t < 2) {
    cout << t << endl;
    return 0;
  }
  for (i = 0; i < 2; i++)
    for (j = i + 1; j < i + 3; j++) {
      long double p = 1. * a[j] / a[i], x = a[j];
      if (a[i] == 0) p = 0;
      int k = j - 1;
      for (int u = j; u < n; u++) {
        if (fabs(x - a[u]) > 1e-4) {
          k++;
          if (k == 2) break;
        } else
          x *= p;
      }
      m = min(m, k);
    }
  cout << m << endl;
  return 0;
}
