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 a[510000];
int n;
bool Dif(int pos) {
  for (int i = 0; i < n; i++) {
    if (i == pos) continue;
    int next1 = i + 1;
    if (next1 == pos) next1++;
    if (next1 >= n) continue;
    if (a[i] == 0 && a[next1] != 0) return false;
    int next2 = next1 + 1;
    if (next2 == pos) next2++;
    if (next2 >= n) break;
    if (a[next1] * a[next1] == a[next2] * a[i])
      continue;
    else
      return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int pos = -1;
  if (a[0] == 0 && a[1] != 0) pos = 0;
  for (int i = 1; i < n - 1 && pos == -1; i++)
    if (a[i] * a[i] != a[i - 1] * a[i + 1] || (a[i] == 0 && a[i + 1] != 0)) {
      pos = i - 1;
      break;
    }
  if (pos == -1) {
    puts("0");
    return 0;
  }
  for (int i = 0; i < 3; i++)
    if (Dif(pos + i) == true) {
      puts("1");
      return 0;
    }
  puts("2");
  return 0;
}
