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;
static const double EPS = 1e-6;
static const double PI = acos(-1.0);
double b, c;
int n;
int seq[110000];
int memo[100010][2][2];
int calc(int depth, int skip, int puse) {
  if (memo[depth][skip][puse] != -1) {
    return memo[depth][skip][puse];
  }
  if (depth >= n) {
    return skip;
  }
  int ret = 2;
  int prev = puse ? depth - 1 : depth - 2;
  if (fabs(seq[depth] - seq[prev] * b) < EPS) {
    ret = min(ret, calc(depth + 1, skip, 1));
  }
  if (!skip) {
    ret = min(ret, calc(depth + 1, 1, 0));
  }
  return memo[depth][skip][puse] = ret;
}
int main() {
  while (scanf("%d", &n) > 0) {
    for (int i = 0; i < (int)(n); i++) {
      scanf("%d", &seq[i]);
    }
    seq[n] = 0;
    int ans = 2;
    if (n <= 1) {
      ans = 0;
    } else {
      c = seq[0];
      if (c != 0 || seq[1] == 0) {
        b = seq[1] / c;
        if (seq[1] == 0) {
          b = 0;
        }
        memset((memo), -1, sizeof(memo));
        ans = min(ans, calc(2, 0, 1));
      }
      c = seq[0];
      if (c != 0 || seq[2] == 0) {
        b = seq[2] / c;
        if (seq[2] == 0) {
          b = 0;
        }
        memset((memo), -1, sizeof(memo));
        ans = min(ans, calc(3, 1, 1));
      }
      c = seq[1];
      if (c != 0 || seq[2] == 0) {
        b = seq[2] / c;
        if (seq[2] == 0) {
          b = 0;
        }
        memset((memo), -1, sizeof(memo));
        ans = min(ans, calc(3, 1, 1));
      }
    }
    printf("%d\n", ans);
  }
}
