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 maxn = 100000 + 10;
const double eps = 1e-6;
int n, a[maxn];
double b;
bool iseq(double a, double b) {
  if (a < b) swap(a, b);
  return a - b < eps;
}
bool check(int ind) {
  if (n == 2) {
    if (ind == -1)
      if (a[0] == 0 && a[1] != 0)
        return false;
      else
        return true;
    return true;
  }
  int tmp;
  if (ind != -1) {
    tmp = a[ind];
    for (int i = ind; i < n - 1; i++) a[i] = a[i + 1];
    n--;
  }
  bool sw = true;
  double c = 0;
  if (a[0] != 0) c = double(a[1]) / a[0];
  for (int i = 1; i < n; i++)
    if (!iseq(a[i - 1] * c, a[i])) sw = false;
  if (ind != -1) {
    for (int i = n; i > ind; i--) a[i] = a[i - 1];
    a[ind] = tmp;
    n++;
  }
  return sw;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (check(-1)) {
    cout << 0 << endl;
    return 0;
  }
  if (check(0) || check(1)) {
    cout << 1 << endl;
    return 0;
  } else {
    b = 0;
    if (a[0] != 0) b = (double)a[1] / a[0];
    for (int i = 1; i < n; i++)
      if (!iseq(a[i - 1] * b, a[i])) {
        if (check(i)) {
          cout << 1 << endl;
          return 0;
        } else {
          cout << 2 << endl;
          return 0;
        }
      }
    cout << 0 << endl;
    return 0;
  }
  cout << 2 << endl;
  return 0;
}
