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 check(vector<int> A) {
  int N = A.size();
  double R = A[0] ? 1. * A[1] / A[0] : 1;
  double a = A[0];
  bool v = 1;
  for (int i = (1); i < (N) && (v); ++i) a *= R, v = (fabs(a - A[i]) < 1e-9);
  return v;
}
int main(int argc, char **argv) {
  int N;
  cin >> N;
  int A[N];
  for (int i = (0); i < (N); ++i) scanf("%d", A + i);
  vector<int> S(N);
  for (int i = (0); i < (N); ++i) S[i] = A[i];
  bool v = check(S);
  if (v) {
    puts("0");
    return 0;
  }
  S.erase(S.begin());
  v = check(S);
  if (v) {
    puts("1");
    return 0;
  }
  S.clear();
  for (int i = (0); i < (N); ++i)
    if (i != 1) S.push_back(A[i]);
  v = check(S);
  if (v) {
    puts("1");
    return 0;
  }
  double R = A[0] ? 1. * A[1] / A[0] : 1;
  double a = A[0];
  v = 1;
  int fix = 0;
  a *= R * R;
  for (int i = (2); i < (N) && (fix < 2); ++i) {
    if (fabs(a - A[i]) >= 1e-9) {
      fix++;
    } else
      a *= R;
  }
  puts((fix < 2) ? "1" : "2");
  return 0;
}
