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>
const int inf = 1000 * 1000 * 1000;
using namespace std;
int a[100005];
int n;
int solve() {
  int il, ir;
  if (n == 1) return 0;
  int zero = 0;
  for (int i = 0; i < (int)(n); i++)
    if (a[i] == 0) zero++;
  if (zero == n || (zero == n - 1 && a[0] != 0)) return 0;
  if (zero == n - 1 || (n != 2 && zero == n - 2 && a[0] != 0)) return 1;
  if (zero > 1) return 2;
  for (il = 1; il <= n - 2; il++)
    if (a[il] * a[il] != a[il - 1] * a[il + 1]) break;
  for (ir = n - 2; ir >= 1; ir--)
    if (a[ir] * a[ir] != a[ir - 1] * a[ir + 1]) break;
  if (il == n - 1) return 0;
  if (il == n - 2 || ir == 1) return 1;
  bool solve = false;
  for (int i = ir - 1; i <= il + 1; i++) {
    bool ok = true;
    if (i > 1 && a[i - 1] * a[i - 1] != a[i - 2] * a[i + 1]) ok = false;
    if (i < n - 2 && a[i + 1] * a[i + 1] != a[i - 1] * a[i + 2]) ok = false;
    if (ok) {
      solve = true;
      break;
    }
  }
  if (solve)
    return 1;
  else
    return 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", a + i);
  printf("%d", solve());
  return 0;
}
