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 = 1000000000;
const int NMAX = 100000;
int N, T[NMAX];
int check(int ifir, int isec) {
  if (T[ifir] == 0) {
    int cnt = 0;
    for (int ind = (isec), _n = (N - 1); ind <= _n; ++ind)
      if (T[ind] != 0) ++cnt;
    return min(cnt, 2);
  }
  int cnt = 0, ilast = isec, inext = ilast + 1;
  while (inext < N) {
    if (T[inext] * T[ifir] == T[ilast] * T[isec]) {
      ilast = inext++;
    } else {
      ++inext;
      if (++cnt == 2) return cnt;
    }
  }
  return cnt;
}
int main() {
  scanf("%d", &N);
  for (int i = 0, _n = (N); i < _n; ++i) scanf("%d", T + i);
  if (N <= 2) {
    printf("%d\n", N == 2 && T[0] == 0 && T[1] != 0 ? 1 : 0);
    return 0;
  }
  int res = check(0, 1);
  if (res <= 1)
    printf("%d\n", res);
  else
    printf("%d\n", min(2, 1 + min(check(0, 2), check(1, 2))));
}
