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 std::max;
using std::min;
const int N = 100000;
int n, a[N], length[N][3];
int solve(int i, int j) {
  if (a[i] == 0 && a[j] != 0) {
    return INT_MIN;
  }
  int &ret = length[i][j - i];
  if (ret == -1) {
    ret = 2;
    for (int k = j + 1; k < n && k < j + 3; ++k) {
      if (a[i] * a[k] == a[j] * a[j]) {
        ret = max(ret, solve(j, k) + 1);
      }
    }
  }
  return ret;
}
int work() {
  memset(length, -1, sizeof(length));
  int maximum = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n && j < i + 3; ++j) {
      maximum = max(maximum, solve(i, j));
    }
  }
  return min(n - maximum, 2);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  printf("%d\n", work());
  return 0;
}
