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[100001];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1) return puts("0");
  if (n == 2 && !a[0] && a[1]) return puts("1");
  if (n == 2) return puts("0");
  int ans = 2;
  for (int i = 0; i < 3; i++)
    for (int j = i + 1; j < 3; j++) {
      if (!a[i] && a[j]) continue;
      double r = a[i] ? 1. * a[j] / a[i] : 0;
      int cnt = (i + j >= 2), last = j;
      for (int k = j + 1; k < n; k++)
        if (abs(a[last] * r - a[k]) < 1e-9)
          last = k;
        else
          cnt++;
      ans = min(ans, cnt);
    }
  printf("%d\n", ans);
}
