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>
template <typename T>
inline T min(const T a, const T b) {
  return a < b ? a : b;
}
template <typename T>
inline T min(const T a, const T b, const T c) {
  return min(a, min(b, c));
}
static const int MAXN = 1e5 + 6;
int n;
int a[MAXN];
int check(int a0, int a1, int p, bool skipped) {
  for (; p < n; ++p) {
    if (!(a0 == 0 && a1 != 0) && !(a1 == 0 && a[p] != 0) &&
        (long long)a[p] * a0 == (long long)a1 * a1) {
      a0 = a1;
      a1 = a[p];
    } else if (!skipped) {
      skipped = true;
    } else
      return 2;
  }
  return (skipped ? 1 : 0);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  if (n <= 2) {
    puts((n == 2 && a[0] == 0 && a[1] != 0) ? "1" : "0");
    return 0;
  }
  printf("%d\n", min(check(a[0], a[1], 2, false), check(a[0], a[2], 3, true),
                     check(a[1], a[2], 3, true)));
  return 0;
}
