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;
int T[100005];
int N;
bool check(vector<int>& V) {
  int a;
  for (a = 0; a < V.size(); a++)
    if (V[a] == 0) break;
  if (a < N && a > 1) return false;
  for (int b = a; b < V.size(); b++)
    if (V[b] != 0) return false;
  if (a < N) return true;
  for (int i = 1; i < V.size() - 1; i++)
    if (V[i] * V[i] != V[i - 1] * V[i + 1]) return false;
  return true;
}
int main() {
  int N;
  scanf("%d", &N);
  int T[N];
  for (int i = 0; i < N; i++) scanf("%d", &T[i]);
  vector<int> v;
  for (int i = 0; i < N; i++) v.push_back(T[i]);
  if (check(v)) {
    printf("0\n");
    return 0;
  }
  v.clear();
  for (int i = 0; i < N; i++)
    if (i != 0) v.push_back(T[i]);
  if (check(v)) {
    printf("1\n");
    return 0;
  }
  v.clear();
  for (int i = 0; i < N; i++)
    if (i != 1) v.push_back(T[i]);
  if (check(v)) {
    printf("1\n");
    return 0;
  }
  int k;
  bool zero = (T[0] == 0) || (T[1] == 0);
  if (zero) {
    for (k = 2;; k++)
      if (T[k] != 0) break;
  } else {
    for (k = 2;; k++)
      if (T[k - 1] * T[1] != T[k] * T[0]) break;
  }
  v.clear();
  for (int i = 0; i < N; i++)
    if (i != k) v.push_back(T[i]);
  if (check(v)) {
    printf("1\n");
    return 0;
  }
  printf("2\n");
  return 0;
}
