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 LIM_N = 1e5 + 5;
int main() {
  int N;
  long long nums[LIM_N];
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%lld", &nums[i]);
  if (N == 1) {
    printf("0\n");
    return 0;
  }
  bool ans[3] = {false, false, true};
  for (int w = 0; w < 2; w++) {
    for (int i = 0; i < 3 && i < N; i++) {
      for (int j = i + 1; j < 3 && j < N; j++) {
        if (nums[i] && nums[j] % nums[i]) continue;
        long long c = nums[i];
        long long b = c ? nums[j] / c : 1;
        if (b == 0 && w) continue;
        int rep = 0;
        for (int k = 0; k < N && rep < 2; k++) {
          if (nums[k] != c) {
            rep++;
            continue;
          }
          c *= b;
        }
        if (rep < 2) ans[rep] = true;
      }
    }
    reverse(nums, nums + N);
  }
  for (int i = 0; i < 3; i++)
    if (ans[i]) {
      printf("%d\n", i);
      return 0;
    }
}
