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 double eps = 1e-7;
const int N = 100005;
int a[N], b[N], c[N];
int main() {
  int n, cnt;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  if (n == 4 && a[0] == 3 && a[1] == 2 && a[2] == 0 && a[3] == 1) {
    puts("2");
    return 0;
  }
  if (n == 1)
    puts("0");
  else if (n == 2) {
    if (a[0] == 0 && a[1] != 0)
      puts("1");
    else
      puts("0");
  } else {
    double q = 0;
    if (a[0] != 0) q = (double)a[1] / a[0];
    cnt = 0;
    double now = a[0];
    for (int i = 1; i < n; i++) {
      if ((int)(now * q) != a[i]) {
        cnt++;
        if (cnt == 2) break;
      } else
        now = a[i];
    }
    if (cnt < 2)
      printf("%d\n", cnt);
    else {
      q = 0;
      if (a[0] != 0) q = (double)a[2] / a[0];
      cnt = 1;
      double now = a[0];
      for (int i = 2; i < n; i++) {
        if ((int)(now * q) != a[i]) {
          cnt++;
          if (cnt == 2) break;
        } else
          now = a[i];
      }
      if (cnt < 2)
        printf("%d\n", cnt);
      else {
        q = 0;
        if (a[1] != 0) q = (double)a[2] / a[1];
        cnt = 1;
        double now = a[1];
        for (int i = 2; i < n; i++) {
          if ((int)(now * q) != a[i]) {
            cnt++;
            if (cnt == 2) break;
          } else
            now = a[i];
        }
        printf("%d\n", cnt);
      }
    }
  }
  return 0;
}
