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>
int n;
int s[100050];
inline bool check(int *s, int n) {
  if (n == 1) return 1;
  if (!s[n - 2] && s[n - 1]) return 0;
  for (int i = 1; i < n - 1; i++)
    if (s[i] * s[i] != s[i - 1] * s[i + 1]) return 0;
  return 1;
}
inline int solve() {
  if (check(s, n)) return 0;
  if (check(s + 1, n - 1)) return 1;
  int t = s[1];
  s[1] = s[0];
  if (check(s + 1, n - 1)) return 1;
  s[1] = t;
  int c = s[1], d = s[0];
  bool flag = 0;
  if (d == 0) {
    if (c) return 2;
    for (int i = 1; i < n; i++) {
      if (s[i]) {
        if (flag) return 2;
        flag = 1;
      }
    }
    return 1;
  }
  int p = s[0];
  for (int i = 1; i < n; i++) {
    if (p * c % d || s[i] != p * c / d) {
      if (flag) return 2;
      flag = 1;
      continue;
    }
    p = s[i];
  }
  return 1;
}
int main(void) {
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) scanf("%d", s + i);
    printf("%d\n", solve());
  }
  return 0;
}
