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], ans = 0;
int calc(int st, int sum) {
  for (int i = st + 2; i <= n; i++) {
    if (a[i] * a[st] != a[st + 1] * a[i - 1]) {
      sum++;
      if (i < n)
        if (a[i + 1] * a[st] == a[i - 1] * a[st + 1])
          i++;
        else
          sum++;
    }
    if (!a[st] && a[i]) sum++;
  }
  return min(sum, 2);
}
void solve() {
  if (n == 1) return;
  if (n == 2) {
    if (!a[1] && a[2]) ans = 1;
    return;
  }
  ans = calc(1, 0);
  if (ans == 0 || ans == 1) return;
  ans = min(2, calc(2, 0) + 1);
  if (ans == 1) return;
  a[2] = a[1], ans = min(2, calc(2, 0) + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve();
  printf("%d\n", ans);
  return 0;
}
