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;
long long int n, arr[123456], s1, s2, s3;
long double c, b, ans = 0.0000001;
long long int check1() {
  long long int x = 0;
  c = arr[0];
  c == 0 ? b = 1 : b = arr[1] / c;
  for (int i = 0; i < n; i++) {
    if (c * pow(b, x) >= arr[i] - ans && c * pow(b, x) <= arr[i] + ans) {
      x++;
    }
    if (i - x >= 2) break;
  }
  return n - x;
}
long long int check2() {
  long long int x = 0;
  c = arr[0];
  c == 0 ? b = 1 : b = arr[2] / c;
  for (int i = 0; i < n; i++) {
    if (c * pow(b, x) >= arr[i] - ans && c * pow(b, x) <= arr[i] + ans) {
      x++;
    }
    if (i - x >= 2) break;
  }
  return n - x;
}
long long int check3() {
  long long int x = 0;
  c = arr[1];
  c == 0 ? b = 1 : b = arr[2] / c;
  for (int i = 0; i < n; i++) {
    if (c * pow(b, x) >= arr[i] - ans && c * pow(b, x) <= arr[i] + ans) {
      x++;
    }
    if (i - x >= 2) break;
  }
  return n - x;
}
int main(int argc, char** argv) {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  s1 = check1();
  s2 = check2();
  s3 = check3();
  cout << min(s1, min(s2, min(s3, (long long int)2)));
  return 0;
}
