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;
int a[100000], z;
int b[100000];
bool solve(int ind) {
  for (int i = 0; i < (int)(n - 1); ++i)
    if (i < ind) {
      b[i] = a[i];
    } else {
      b[i] = a[i + 1];
    }
  for (int i = 0; i < (int)(n - 3); ++i)
    if (b[i + 2] * b[i] != b[i + 1] * b[i + 1]) {
      return false;
    }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); ++i) scanf("%d", a + i), z += a[i] == 0;
  if (z) {
    int nz = 0;
    for (int i = 1; i < n; ++i) {
      nz += a[i] != 0;
    }
    if (nz > 2) nz = 2;
    if (z == 1) {
      for (int i = 0; i < (int)(n - 1); ++i)
        if (a[i] == 0) {
          swap(a[i], a[i + 1]);
        }
      --n;
      int cur = 1;
      for (int i = 0; i < (int)(n - 2); ++i)
        if (a[i + 2] * a[i] != a[i + 1] * a[i + 1]) {
          cur = 2;
          break;
        }
      nz = min(nz, cur);
    }
    printf("%d\n", nz);
    return 0;
  }
  for (int i = 0; i < (int)(n - 2); ++i)
    if (a[i] * a[i + 2] != a[i + 1] * a[i + 1]) {
      if (solve(i) || solve(i + 1) || solve(i + 2)) {
        printf("1\n");
      } else {
        printf("2\n");
      }
      return 0;
    }
  printf("0\n");
  return 0;
}
