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 org[100000 + 100];
long long num[100000 + 100];
int check(long long x[], int n) {
  int i;
  if (n <= 2) return 0;
  for (i = 3; i <= n; i++)
    if (x[i] * x[1] != x[2] * x[i - 1]) return i;
  return 0;
}
int main() {
  int n;
  int i, temp, zero = 0, pos;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> org[i];
    if (org[i] == 0) {
      pos = i;
      zero++;
    }
  }
  if (zero) {
    int p;
    if (zero == n)
      printf("0\n");
    else if (zero == n - 1) {
      if (org[1] != 0)
        printf("0\n");
      else
        printf("1\n");
    } else if (zero == n - 2 && org[1] != 0)
      printf("1\n");
    else if (zero == 1) {
      p = 0;
      for (i = 1; i <= n; i++)
        if (i != pos) num[++p] = org[i];
      if (check(num, n - 1) == 0)
        printf("1\n");
      else
        printf("2\n");
    } else
      printf("2\n");
    return 0;
  }
  temp = check(org, n);
  if (temp == 0)
    printf("0\n");
  else if (temp > 3) {
    int p = 0;
    int res;
    for (i = 1; i <= n; i++)
      if (i != temp) num[++p] = org[i];
    res = check(num, n - 1);
    if (res == 0)
      printf("1\n");
    else
      printf("2\n");
  } else {
    int p = 0;
    int res;
    for (i = 1; i <= n; i++)
      if (i != 1) num[++p] = org[i];
    if (check(num, n - 1) == 0) {
      printf("1\n");
      return 0;
    }
    p = 0;
    for (i = 1; i <= n; i++)
      if (i != 2) num[++p] = org[i];
    if (check(num, n - 1) == 0) {
      printf("1\n");
      return 0;
    }
    p = 0;
    for (i = 1; i <= n; i++)
      if (i != 3) num[++p] = org[i];
    if (check(num, n - 1) == 0) {
      printf("1\n");
      return 0;
    }
    printf("2\n");
    return 0;
  }
  return 0;
}
