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 n, t[100010];
double b;
int check0() {
  if (!t[0]) {
    int ile = 0;
    for (int i = 1; i < n; i++)
      if (t[i] != 0) ile++;
    if (ile < 3) return ile;
    return 2;
  }
  if (n == 1 || n == 2) return 0;
  b = t[1] / (double)t[0];
  int pos = -1;
  double pom = t[1];
  for (int i = 2; i < n; i++) {
    pom *= b;
    if (pom != t[i]) {
      pos = i;
      break;
    }
  }
  if (pos == -1) return 0;
  int pos1 = -1;
  pom = t[0];
  for (int i = 1; i < n; i++) {
    if (i == pos) continue;
    pom *= b;
    if (pom != t[i]) {
      pos1 = i;
      break;
    }
  }
  if (pos1 == -1) return 1;
  if (!t[pos]) return 2;
  b = pow(t[pos] / (double)t[0], 1 / (double)(pos - 1));
  int ile = 0;
  pom = t[0];
  for (int i = 1; i < n; i++) {
    pom *= b;
    if (t[i] != pom) {
      ile++;
      pom /= b;
    }
  }
  if (ile == 1) return 1;
  return 2;
}
int check1() {
  if (n - 1 < 3) return 1;
  b = t[2] / (double)t[1];
  double pom = t[2];
  for (int i = 3; i < n; i++) {
    pom *= b;
    if (t[i] != pom) return 2;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &t[i]);
  printf("%d\n", min(check0(), check1()));
  return 0;
}
