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;
inline void pisz(int n) { printf("%d\n", n); }
vector<int> v;
int n;
int checkwyw(vector<int> &a) {
  if (a.size() <= 2) return true;
  int f = a[0], s = a[1];
  bool ok = true;
  for (int i = 2; i < a.size(); ++i) {
    if (f * a[i] != s * a[i - 1]) {
      ok = false;
      break;
    }
  }
  return ok;
}
int check(vector<int> w) {
  if (w.size() <= 2) return 0;
  if (checkwyw(w)) return 0;
  vector<int> ww;
  for (int(i) = 0; (i) < (w.size()); ++(i))
    if (i != 0) ww.push_back(w[i]);
  if (checkwyw(ww)) return 1;
  ww.clear();
  for (int(i) = 0; (i) < (w.size()); ++(i))
    if (i != 1) ww.push_back(w[i]);
  if (checkwyw(ww)) return 1;
  int f = w[0], s = w[1];
  for (int i = 2; i < w.size(); ++i) {
    if (f * w[i] != s * w[i - 1]) {
      ww.clear();
      for (int(k) = 0; (k) < (w.size()); ++(k))
        if (k != i) ww.push_back(w[k]);
      if (checkwyw(ww))
        return 1;
      else
        return 2;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); ++(i)) {
    int(fdg);
    scanf("%d", &(fdg));
    ;
    v.push_back(fdg);
  }
  int zero = 0;
  for (int(i) = 0; (i) < (n); ++(i))
    if (v[i] == 0) ++zero;
  if (zero) {
    int nonzero = 0;
    for (int(i) = 0; (i) < (n); ++(i))
      if (v[i] != 0) ++nonzero;
    if (nonzero == 0) {
      printf("0");
      return 0;
    }
    if (nonzero == 1) {
      if (v[0] != 0)
        printf("0");
      else
        printf("1");
      return 0;
    }
    if (zero == 1) {
      vector<int> w;
      for (int(i) = 0; (i) < (n); ++(i))
        if (v[i] != 0) w.push_back(v[i]);
      bool dasie = checkwyw(w);
      if (dasie)
        printf("1");
      else
        printf("2");
      return 0;
    }
    if (nonzero == 2) {
      if (v[0] != 0)
        printf("1");
      else
        printf("2");
      return 0;
    }
    printf("2");
    return 0;
  }
  pisz(check(v));
  return 0;
}
