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 a[100000], b[100000];
bool check(int n) {
  for (int i = 1; i + 1 < n; i++)
    if (a[i] * a[i] != a[i - 1] * a[i + 1]) return 0;
  if (n > 1 && !a[n - 2] && a[n - 1]) return 0;
  return 1;
}
bool check2(int n) {
  for (int i = 0; i < n; i++) b[i] = a[i];
  for (int i = 1; i + 1 < n; i++)
    if (b[i] * b[i] != b[i - 1] * b[i + 1]) {
      for (int j = 0; j <= i; j++) a[j] = b[j];
      for (int j = i + 2; j < n; j++) a[j - 1] = b[j];
      if (check(n - 1)) return 1;
      for (int j = 0; j < i; j++) a[j] = b[j];
      for (int j = i + 1; j < n; j++) a[j - 1] = b[j];
      if (check(n - 1)) return 1;
      for (int j = 0; j < i - 1; j++) a[j] = b[j];
      for (int j = i; j < n; j++) a[j - 1] = b[j];
      if (check(n - 1)) return 1;
      return 0;
    }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (check(n))
    puts("0");
  else if (check2(n))
    puts("1");
  else
    puts("2");
  return 0;
}
