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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
int a, N, A[100005];
int gp(int N, int *A) {
  if (N == 1) return 0;
  int z = 0, S = 0;
  for (int i(0); i < N; i++) z += !A[i];
  if (!*A) return N - z;
  double g = A[1] / (double)*A, l = A[1];
  for (int k(2); k < N; k++)
    if ((l * g + (1e-10) > A[k] && l * g - (1e-10) < A[k]))
      l = A[k];
    else
      ++S;
  return S;
}
int main(void) {
  scanf("%d", &N), ga(N, A), a = gp(N, A);
  if (a < 2) return printf("%d\n", a), 0;
  a = gp(N - 1, A + 1);
  if (!a) return puts("1");
  swap(A[1], *A), a = gp(N - 1, A + 1);
  if (!a) return puts("1");
  puts("2");
  return 0;
}
