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 INF = 1000000007;
long double divide(int a, int b) {
  if (a == 0) return 0;
  if (b == 0) return INF;
  return (1.0 * a) / b;
}
int niz[100005];
int novi[100005];
long double raz[100005];
map<long double, int> br;
int main() {
  ios_base::sync_with_stdio(false);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> niz[i];
  }
  if (n == 1) {
    cout << 0;
    return 0;
  }
  if (n == 2) {
    if (niz[1] == 0 && niz[2] != 0) {
      cout << 1;
    } else
      cout << 0;
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    raz[i] = divide(niz[i], niz[i - 1]);
    br[raz[i]]++;
  }
  if (br[raz[2]] == n - 1) {
    cout << 0;
    return 0;
  }
  if (n <= 10) {
    for (int i = 1; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        novi[j - 1] = niz[j];
      }
      long double r1 = divide(novi[2], novi[1]);
      for (int j = 2; j < n; j++) {
        long double r2 = divide(novi[j], novi[j - 1]);
        if (r1 != r2) break;
        if (j == n - 1) {
          cout << 1;
          return 0;
        }
      }
      novi[i] = niz[i];
    }
    cout << 2;
    return 0;
  }
  if (br[raz[3]] == n - 2) {
    if (!(raz[2] == raz[3])) {
      cout << 1;
      return 0;
    }
    if (!(raz[n] == raz[3])) {
      cout << 1;
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    if (br[raz[i]] >= n - 3) {
      for (int j = 2; j < n; j++) {
        if (!(raz[j] == raz[i])) {
          long double a = divide(niz[j + 1], niz[j - 1]);
          if (a == raz[i]) {
            cout << 1;
            return 0;
          }
        }
      }
    }
  }
  cout << 2;
  return 0;
}
