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, a[100010];
bool test(int i, int j, int k) {
  if (a[i] == 0) return a[j] == 0 && a[k] == 0;
  return a[j] * a[j] == a[k] * a[i];
}
int solve() {
  bool ok = true;
  for (int i = 0; i < (int)(n); i++)
    if (a[i] != a[0]) {
      ok = false;
      break;
    }
  if (ok) return 0;
  if (n == 2)
    if (a[0] != 0)
      return 0;
    else
      return 1;
  int i = 1;
  while (i < n - 1 && test(i - 1, i, i + 1)) i++;
  int j = n - 2;
  while (j > 0 && test(j - 1, j, j + 1)) j--;
  if (i == n - 1) return 0;
  if (i == n - 2 || j == 1) return 1;
  if (i + 2 == j && test(i - 1, i, j) && test(i, j, j + 1)) return 1;
  if (i + 1 == j && test(i - 1, i, j + 1) &&
      (j + 2 >= n || test(i, j + 1, j + 2)))
    return 1;
  if (i + 1 == j && (i - 2 < 0 || test(i - 2, i - 1, j)) &&
      test(i - 1, j, j + 1))
    return 1;
  return 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  printf("%d\n", solve());
}
