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[120000], ans, i, n;
void check(double x, double y) {
  int i, cnt = 0;
  for (i = 1; i <= n; ++i) {
    if (abs(x - a[i]) > 1e-8)
      cnt++;
    else
      x *= y;
  }
  ans = min(ans, cnt);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  if (n == 1) return puts("0"), 0;
  if (n == 2) {
    if (a[1] == 0 && a[2] != 0)
      puts("1");
    else
      puts("0");
    return 0;
  }
  ans = 2;
  check(a[2], !a[2] ? 0 : 1.0 * a[3] / a[2]);
  check(a[1], !a[1] ? 0 : 1.0 * a[3] / a[1]);
  check(a[1], !a[1] ? 0 : 1.0 * a[2] / a[1]);
  printf("%d\n", ans);
}
