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;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
int n, ans = 0x3f3f3f3f, zero;
vi a;
int check(int s, int r, int cnt) {
  if (!r) return 0x3f3f3f3f;
  for (int i = s; i < n - 1; i++)
    if (a[i + 1] != a[i] * r) cnt++, i++;
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (!a[i]) zero++;
  }
  if (n == 2) {
    if (!a[1])
      cout << 0;
    else if (!a[0])
      cout << 1;
    else if (!(a[0] % a[1]) || !(a[1] % a[0]))
      cout << 0;
    else
      cout << 1;
    return 0;
  }
  if (a[0])
    ans = min({ans, check(1, a[1] / a[0], 0), check(2, a[2] / a[0], 1)});
  if (a[1]) ans = min(ans, check(2, a[2] / a[1], 1));
  if (a[0]) ans = min(ans, n - zero - 1);
  reverse((a).begin(), (a).end());
  if (a[0])
    ans = min({ans, check(1, a[1] / a[0], 0), check(2, a[2] / a[0], 1)});
  if (a[1]) ans = min(ans, check(2, a[2] / a[1], 1));
  ans = min(ans, n - zero);
  cout << min(ans, 2);
}
