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 in[100010];
int min(int a, int b) { return a > b ? b : a; }
int main() {
  int i, j, k, n, p, ans;
  double d;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &in[i]);
  }
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  if (n == 2) {
    if (!in[1] && in[2])
      printf("1\n");
    else
      printf("0\n");
    return 0;
  }
  ans = 2;
  for (i = 1; i <= 2; i++)
    for (j = i + 1; j <= 3; j++) {
      if (!in[i] && in[j]) continue;
      if (in[i] == 0)
        d = 0;
      else
        d = (double)in[j] / in[i];
      p = j;
      int s = 0;
      if (i != 1 || j != 2) s++;
      for (k = j + 1; k <= n; k++)
        if (fabs(in[p] * d - in[k]) < 1e-9)
          p = k;
        else
          s++;
      ans = min(ans, s);
    }
  printf("%d\n", ans);
  return 0;
}
