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 ans = 2, n, a[111111], p;
long double d;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1) {
    puts("0");
    return 0;
  } else if (n == 2) {
    if (a[0] == 0 && a[1])
      puts("1");
    else
      puts("0");
    return 0;
  }
  for (int i1 = 0; i1 < 2; i1++)
    for (int i2 = i1 + 1; i2 < 3; i2++) {
      if (a[i1] == 0 && a[i2] != 0) continue;
      if (a[i1] == 0)
        d = 0;
      else
        d = a[i2] / (long double)a[i1];
      p = i2;
      int s = 0;
      if (i1 != 0 || i2 != 1) s++;
      for (int i = i2 + 1; i < n; i++)
        if (fabs(a[p] * d - a[i]) < 1e-9)
          p = i;
        else
          s++;
      ans = min(ans, s);
    }
  cout << ans << endl;
  return 0;
}
