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, i, j, k, ok;
int v[100010], pa[100010], pb[100010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
  pa[1] = pa[2] = 1;
  if (v[1] == 0 && v[2] != 0) pa[2] = 0;
  for (int i = 3; i <= n; ++i)
    if (v[i] * v[i - 2] == v[i - 1] * v[i - 1] && pa[i - 1] == 1 &&
        !(v[i - 1] == 0 && v[i] != 0))
      pa[i] = 1;
    else
      break;
  pb[n] = pb[n - 1] = 1;
  for (int i = n - 2; i; --i)
    if (v[i] * v[i + 2] == v[i + 1] * v[i + 1] &&
        !(v[i + 2] != 0 && v[i + 1] == 0))
      pb[i] = 1;
    else
      break;
  if (pa[n] == 1) {
    printf("0\n");
    return 0;
  }
  if (pb[2] == 1 || pa[n - 1] == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 2; i < n; ++i) {
    if (pa[i - 1] == 1 && pb[i + 1] == 1) {
      ok = 1;
      if (i > 2 && !(v[i - 2] * v[i + 1] == v[i - 1] * v[i - 1] &&
                     !(v[i - 2] == 0 && v[i - 1] != 0)))
        ok = 0;
      if (i < n - 1 && !(v[i + 1] * v[i + 1] == v[i - 1] * v[i + 2] &&
                         !(v[i + 1] == 0 && v[i + 2] != 0)))
        ok = 0;
      if (v[i - 1] == 0 && v[i + 1] != 0) ok = 0;
      if (i > 2 && i < n - 1 && v[i - 2] * v[i + 2] != v[i - 1] * v[i + 1])
        ok = 0;
      if (ok) {
        printf("1\n");
        return 0;
      }
    }
  }
  printf("2\n");
  return 0;
}
