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 Maxn = 100000 + 10;
const int INF = 0x7f7f7f7f;
const double eps = 1e-8;
const double pi = 3.1415926535897932384626433832795;
int a[Maxn];
int main() {
  double d;
  int n, ans = 2;
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n == 1)
    cout << 0 << endl;
  else if (n == 2) {
    if (a[1] == 0 && a[2])
      cout << 1 << endl;
    else
      cout << 0 << endl;
  } else {
    for (int i = 1; i <= 2; i++)
      for (int j = i + 1; j <= 3; j++) {
        if (a[i] == 0 && a[j]) continue;
        if (a[i] == 0)
          d = 0;
        else
          d = 1.0 * a[j] / a[i];
        int p = j, cnt = 0;
        if (i != 1 || j != 2) cnt++;
        for (int k = j + 1; k <= n; k++) {
          if (abs(a[p] * d - a[k]) < eps)
            p = k;
          else
            cnt++;
        }
        ans = min(ans, cnt);
      }
    cout << ans << endl;
  }
  return 0;
}
