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[100005], n;
int check(int A, int B, int pre, int pos, int x) {
  if (!A || !B) return 0;
  int ct = 0;
  for (int i = pos; i < n; i++) {
    if (B * pre != A * a[i] || !a[i]) {
      ct++;
    } else
      pre = a[i];
  }
  if (ct <= x) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int ct0 = 0;
  for (int i = 0; i < n; i++)
    if (!a[i]) ct0++;
  if (ct0 == n || (ct0 == n - 1 && a[0] != 0)) {
    puts("0");
    return 0;
  } else if (ct0) {
    if (ct0 == n - 1 || (ct0 == n - 2 && a[0] != 0)) {
      puts("1");
      return 0;
    }
  }
  int ok = 1;
  for (int i = 0; i < n - 2; i++)
    if (a[i] * a[i + 2] != a[i + 1] * a[i + 1]) ok = 0;
  if (ct0) ok = 0;
  if (ok) {
    puts("0");
    return 0;
  }
  if (check(a[0], a[1], a[1], 2, 1) || check(a[0], a[2], a[2], 3, 0) ||
      check(a[1], a[2], a[2], 3, 0))
    ok = 1;
  puts(ok ? "1" : "2");
  return 0;
}
