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 i, j, n, c, r, a[111111];
long long p, q;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  for (i = 0; i < n; i++)
    if (a[i] == 0) {
      if (i < 3) {
        for (j = 1; j < n; j++)
          if (a[j] == 0) r++;
        if (r == n - 1) {
          puts("0");
          return 0;
        }
        if (r == n - 2) {
          puts("1");
          return 0;
        }
      }
      for (j = r = 0; j < n; j++)
        if (a[j] == 0) {
          r++;
          c = j;
        } else if (j < n - 1 && a[j + 1] != 0) {
          p = a[j + 1];
          q = a[j];
        }
      for (j = 1; j < n; j++)
        if (a[j] != 0 && a[j - 1] != 0 && a[j] * q != p * a[j - 1]) {
          puts("2");
          return 0;
        }
      if (r == 1 && (c == n - 1 || c == 0 || a[c - 1] * p == a[c + 1] * q))
        puts("1");
      else
        puts("2");
      return 0;
    }
  r = 0;
  p = a[1];
  q = a[0];
  for (i = 2; i < n; i++)
    if (a[i] * a[0] != a[i - 1] * a[1]) {
      r++;
      c = i++;
    }
  if (r == 0) {
    puts("0");
    return 0;
  }
  if (r == 1 && (c == n - 1 || a[c - 1] * p == a[c + 1] * q)) {
    puts("1");
    return 0;
  }
  for (r = 0, i = 3; i < n; i++)
    if (a[i] * a[1] != a[i - 1] * a[2]) r++;
  if (r == 0)
    puts("1");
  else
    puts("2");
  return 0;
}
