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>
int d[100010];
int temp[110];
bool check(int n) {
  bool ck = true;
  int i;
  for (i = 0; i < n; ++i)
    if (temp[i] == 0) break;
  if (i < n) {
    for (i = 1; i < n; ++i)
      if (temp[i] != 0) break;
    return i >= n;
  }
  if (n <= 2) return true;
  i = 1;
  while (i < n && temp[i] == 0) i++;
  if (i == n) return true;
  for (i = 1; i < n - 1; ++i) {
    if (temp[i] * temp[i] != temp[i - 1] * temp[i + 1]) ck = false;
  }
  return ck;
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int i, j;
    int z = 0;
    for (i = 0; i < n; ++i) {
      scanf("%d", d + i);
      z += d[i] == 0 ? 1 : 0;
    }
    if (n < 100) {
      for (i = 0; i < n; ++i) temp[i] = d[i];
      if (check(n)) {
        puts("0");
        continue;
      }
      for (i = 0; i < n; ++i) {
        int ct = 0;
        for (int j = 0; j < n; ++j)
          if (j != i) temp[ct++] = d[j];
        if (check(ct)) break;
      }
      if (i < n)
        puts("1");
      else
        puts("2");
    } else {
      if (z == n - 1) {
        puts(d[0] != 0 ? "0" : "1");
        continue;
      }
      if (z == n - 2) {
        puts(d[0] != 0 ? "1" : "2");
        continue;
      }
      int ct[2], b = -1;
      for (i = 0; i < 2; ++i) {
        ct[i] = 0;
        b *= -1;
        for (j = 1; j < n; ++j) {
          if (d[j - 1] != b * d[j]) ct[i]++;
        }
      }
      if (ct[0] == 0 || ct[1] == 0)
        puts("0");
      else if (ct[0] > 2 && ct[1] > 2)
        puts("2");
      else
        puts("1");
    }
  }
  return 0;
}
