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;
bool isGP3(int a, int b, int c) {
  if (b != 0)
    return (b * b == a * c);
  else
    return (c == 0);
}
int isGP(int N) {
  int ret = 0;
  if (N <= 1) {
    return 0;
  }
  if (N == 2) {
    int a, b;
    cin >> a;
    cin >> b;
    if (a != 0 && b != 0)
      return 0;
    else if (a == 0 && b == 0)
      return 0;
    else if (a != 0 && b == 0)
      return 0;
    else
      return 1;
  }
  int a, b, c;
  cin >> a;
  cin >> b;
  cin >> c;
  for (int i = 3; i <= N; i++) {
    if (isGP3(a, b, c)) {
      if (i < N) {
        a = b;
        b = c;
        cin >> c;
      }
      continue;
    } else {
      if (ret == 0) {
        if (i < N) {
          int d;
          cin >> d;
          if (isGP3(a, b, d)) {
            ret = 1;
            c = d;
          } else if (isGP3(a, c, d)) {
            ret = 1;
            b = c;
            c = d;
          } else if (isGP3(b, c, d)) {
            ret = 1;
            a = b;
            b = c;
            c = d;
          } else {
            return 2;
          }
        } else {
          ret = 1;
        }
      } else if (ret == 1) {
        return 2;
      }
    }
  }
  return ret;
}
int main() {
  int N;
  cin >> N;
  cout << isGP(N) << endl;
  return 0;
}
