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;
bool isgeo(int p, int q, int r) {
  return q * q == p * r && !((p == 0 || q == 0) && r != 0);
}
int n, a[100000];
void solve() {
  if (n == 1) {
    printf("0\n");
    return;
  }
  if (n == 2) {
    if (a[0] == 0 && a[1] != 0)
      printf("1\n");
    else
      printf("0\n");
    return;
  }
  for (int i = (1); i < (int)(n - 1); i++)
    if (!isgeo(a[i - 1], a[i], a[i + 1])) goto N1;
  printf("0\n");
  return;
N1:
  for (int i = (2); i < (int)(n - 1); i++)
    if (!isgeo(a[i - 1], a[i], a[i + 1])) goto N2;
  printf("1\n");
  return;
N2:
  if (!isgeo(a[0], a[2], a[3])) goto N3;
  for (int i = (3); i < (int)(n - 1); i++)
    if (!isgeo(a[i - 1], a[i], a[i + 1])) goto N3;
  printf("1\n");
  return;
N3:
  int t = 0;
  for (int i = (1); i < (int)(n - 1); i++) {
    if (!isgeo(a[i - 1], a[i], a[i + 1])) {
      if (t) goto N4;
      if (i + 2 < n && !isgeo(a[i - 1], a[i], a[i + 2])) goto N4;
      if (i + 3 < n && !isgeo(a[i], a[i + 2], a[i + 3])) goto N4;
      t++;
      i += 2;
    }
  }
  printf("1\n");
  return;
N4:
  printf("2\n");
}
int main() {
  while (cin >> n) {
    for (int i = 0; i < (int)(n); i++) cin >> a[i];
    solve();
  }
  return 0;
}
