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 int L = 1e5;
int a[L];
int c[L];
int main() {
  int N;
  scanf("%d", &N);
  int cz = 0;
  for (int i = 0; i < N; i++) {
    scanf("%d", a + i);
    if (a[i] == 0) {
      cz++;
    }
  }
  if (a[0] == 0) {
    if (cz == N) {
      puts("0");
      return 0;
    }
  } else {
    int phase = 0;
    for (int i = 2; i < N; i++) {
      if (a[i] * a[0] != a[i - 1] * a[1]) {
        phase++;
        if (i + 1 < N && a[i + 1] * a[0] != a[i - 1] * a[1]) {
          phase++;
        } else {
          i++;
        }
      }
    }
    if (phase < 2) {
      printf("%d\n", phase);
      return 0;
    }
  }
  if (cz == N - 1) {
    puts("1");
    return 0;
  }
  for (int k = 0; k < 2; k++) {
    c[0] = a[k];
    if (c[0] == 0) continue;
    for (int i = 2; i < N; i++) {
      c[i - 1] = a[i];
    }
    bool ok = true;
    for (int i = 2; i < N - 1; i++) {
      if (c[i] * c[0] != c[i - 1] * c[1]) {
        ok = false;
      }
    }
    if (ok) {
      puts("1");
      return 0;
    }
  }
  puts("2");
  return 0;
}
