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;
double a[110000], b[110000];
int n;
int check(double a1, double a2) {
  int i, cnt = 0;
  for (i = 1; i <= n; i++)
    if (fabs(a[i] - a1) < 1e-6) {
      cnt++;
      a1 *= a2;
    }
  if (cnt == n) return 0;
  if (cnt == n - 1) return 1;
  return 2;
}
int main() {
  int i, ans = 2, tot = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lf", &a[i]);
  for (i = 1; i <= n; i++)
    if (!(fabs(a[i]) < 1e-6)) b[++tot] = a[i];
  ans = min(check(0, 0), check(b[1], 0));
  if (tot >= 1) ans = min(ans, check(b[1], 1));
  if (tot >= 2) ans = min(ans, check(b[1], b[2] / b[1]));
  if (tot >= 3)
    ans = min(ans, min(check(b[1], b[3] / b[1]), check(b[2], b[3] / b[2])));
  printf("%d\n", ans);
  return 0;
}
