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 a[100010];
int n;
int check1(long long numer, long long denom) {
  if (numer == 0 && denom == 0) {
    int count = 0;
    for (int i = 0; i < n; i++)
      if (a[i] != 0) count++;
    if (count >= 2)
      return 2;
    else
      return count;
  }
  int count = 0;
  int prev = 0;
  for (int i = 1; i < n; i++) {
    if (a[prev] * numer != a[i] * denom)
      count++;
    else
      prev = i;
    if (count > 1) return 2;
  }
  return count;
}
int check2(long long numer, long long denom) {
  if (numer == 0 && denom == 0) {
    int count = 1;
    for (int i = 1; i < n; i++)
      if (a[i] != 0) count++;
    if (count >= 2)
      return 2;
    else
      return count;
  }
  int count = 1;
  int prev = 1;
  for (int i = 2; i < n; i++) {
    if (a[prev] * numer != a[i] * denom)
      count++;
    else
      prev = i;
    if (count > 1) return 2;
  }
  return count;
}
int solve() {
  if (n == 1) return 0;
  if (n == 2) {
    if (a[0] == 0 && a[1] != 0)
      return 1;
    else
      return 0;
  }
  int c1, c2, c3;
  c1 = c2 = c3 = 2;
  if (!(a[0] == 0 && a[1] != 0)) {
    long long numer = a[1];
    long long denom = a[0];
    c1 = check1(numer, denom);
  }
  if (!(a[0] == 0 && a[2] != 0)) {
    long long numer = a[2];
    long long denom = a[0];
    c2 = check1(numer, denom);
  }
  if (!(a[1] == 0 && a[2] != 0)) {
    long long numer = a[2];
    long long denom = a[1];
    c3 = check2(numer, denom);
  }
  return min(c1, min(c2, c3));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  printf("%d\n", solve());
}
