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 N = int(2e5);
int n, a[N], bn, b[N];
int mn, pn, k;
void print(char ch) {
  cout << ch;
  exit(0);
}
int solve(int k) {
  bn = 0;
  for (int i = 1; i <= n; i++) {
    if (i != k) b[++bn] = a[i];
  }
  for (int i = 2; i < bn; i++) {
    if (b[i] * b[i] != b[i - 1] * b[i + 1]) return 2;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!a[i]) k++;
  }
  if (k > 1) {
    if (k == n) print('0');
    if (k == n - 1 && a[1]) print('0');
    if (k == n - 1) print('1');
    if (k == n - 2 && a[1]) print('1');
    print('2');
  }
  for (int i = 2; i < n; i++) {
    if (a[i] * a[i] != a[i - 1] * a[i + 1]) {
      k = i;
      break;
    }
  }
  if (n == 2) {
    if (a[1] == a[2]) print('0');
    if (a[1] == 0) print('1');
    print('0');
  }
  if (solve(-1) != 2) print('0');
  cout << min(solve(k), min(solve(k - 1), solve(k + 1)));
  return 0;
}
