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;
const int inf = 1000 * 1000 * 1000;
int n = 0;
double ar[100009];
int f(double a1, double a2, int bad, int pos) {
  double c = a1, b;
  if (c == 0 && a2 != 0) return 2;
  if (c == 0 && a2 == 0)
    b = 0;
  else
    b = a2 / c;
  int now = a2;
  for (int i = pos; i < n; i++) {
    if (ar[i] == now * b) {
      now *= b;
    } else
      bad++;
  }
  return bad;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lf", ar + i);
  int res = 2;
  res = min(res, f(ar[0], ar[1], 0, 2));
  res = min(res, f(ar[0], ar[2], 1, 3));
  res = min(res, f(ar[1], ar[2], 1, 3));
  printf("%d\n", res);
  return 0;
}
