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;
const int inf = 1999999999;
const double pi = acos(-1.0);
const double eps = 1e-10;
char gc() {
  char c;
  while (isspace(c = getchar()))
    ;
  return c;
}
vector<int> v;
int test(int a, int b, int i) {
  if (a == 0 && b != 0) return 99999;
  int c = 0;
  int cur = b;
  bool update = true;
  while (i < ((int)(v).size())) {
    if (update && a != 0) {
      cur *= b;
      if ((cur % a) != 0) {
        return c + ((int)(v).size()) - i;
      }
      cur /= a;
      if (cur < -10000 || cur > 10000) {
        return c + ((int)(v).size()) - i;
      }
    }
    if (v[i] != cur) {
      c++;
      update = false;
    } else
      update = true;
    i++;
  }
  return c;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    v.push_back(t);
  }
  if (((int)(v).size()) == 1) {
    printf("0");
    return 0;
  }
  if (((int)(v).size()) == 2) {
    if (v[0] != 0)
      printf("0");
    else if (v[1] == 0)
      printf("0");
    else
      printf("1");
    return 0;
  }
  int t = test(v[0], v[1], 2);
  if (t == 0) {
    printf("0");
    return 0;
  }
  if (t == 1) {
    printf("1");
    return 0;
  }
  if (test(v[0], v[2], 3) == 0) {
    printf("1");
    return 0;
  }
  if (test(v[1], v[2], 3) == 0) {
    printf("1");
    return 0;
  }
  printf("2");
}
