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 double Eps = 1e-6;
int n, ans;
double data[100010];
int min(int a, int b) { return a < b ? a : b; }
int check(double a, double q) {
  int t = 0, i;
  for (i = 1; i <= n; i++)
    if (fabs(data[i] - a) < Eps) {
      a = data[i] * q;
      t++;
    }
  return min(n - t, 2);
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lf", &data[i]);
  ans = min(check(0, 0), check(data[1], 0));
  if (n >= 1) ans = min(ans, check(data[1], 1));
  if (n >= 2) ans = min(ans, check(data[1], data[2] / data[1]));
  if (n >= 3)
    ans = min(ans, min(check(data[1], data[3] / data[1]),
                       check(data[2], data[3] / data[2])));
  printf("%d\n", ans);
}
