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 a[100001], n;
bool func(double val, int &k) {
  double x = a[k == 0];
  for (int i = (k != -1) + 1; i < n; i++) {
    if (fabs(x * val - a[i]) < 1e-6)
      x = a[i];
    else if (k != -1)
      return false;
    else
      k = i;
  }
  return true;
}
double findb(double x, double y) {
  if (fabs(x) < 0.5 || fabs(y) < 0.5)
    return 0;
  else
    return y / x;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n == 1) {
    cout << "0\n";
    return 0;
  }
  double b;
  int k = -1;
  b = findb(a[0], a[1]);
  if (func(b, k) == true) {
    if (k == -1)
      cout << "0\n";
    else
      cout << "1\n";
    return 0;
  }
  if (n > 2) {
    k = 1;
    b = findb(a[0], a[2]);
    if (func(b, k) == true) {
      cout << "1\n";
      return 0;
    }
    k = 0;
    b = findb(a[1], a[2]);
    if (func(b, k) == true) {
      cout << "1\n";
      return 0;
    }
  }
  cout << "2\n";
  return 0;
}
