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;
int arr[100010];
int gd(int str) {
  int cnt = 0;
  if (arr[str] == 0) {
    for (int i = str + 1; i < n; i++)
      if (arr[i] != 0) cnt++;
    return min(cnt, 2);
  }
  for (int i = str + 1; i + 1 < n && cnt < 2; i++) {
    if (arr[str] * arr[i + 1] != arr[str + 1] * arr[i]) {
      cnt++;
      if (i + 2 < n && arr[str] * arr[i + 2] != arr[str + 1] * arr[i]) cnt++;
      i++;
    }
  }
  return min(cnt, 2);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  int t1 = gd(0);
  int t2 = gd(1);
  arr[1] = arr[0];
  int t3 = gd(1);
  if (t1 == 0)
    cout << 0 << endl;
  else if (t1 == 1 || t2 == 0 || t3 == 0)
    cout << 1 << endl;
  else
    cout << 2 << endl;
  return 0;
}
