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 INF = 0x3f3f3f3f, N = 2e6 + 5, MOD = 1e9 + 7;
const double eps = 1e-8;
int sgn(double x) { return x < eps ? (x < -eps ? -1 : 0) : 1; }
int n;
int a[N];
int check(double k) {
  int suc = 1;
  for (int i = 2; i <= n; i++)
    if (sgn(k - (double)a[i] / a[i - 1]) != 0) suc = 0;
  if (suc) return 0;
  suc = 1;
  for (int i = 3; i <= n; i++)
    if (sgn(k - (double)a[i] / a[i - 1]) != 0) suc = 0;
  if (suc) return 1;
  int pre, cnt = 0;
  if (sgn(k - (double)a[2] / a[1]) != 0)
    pre = a[1], cnt++;
  else
    pre = a[2];
  for (int i = 3; i <= n; i++)
    if (sgn(k - (double)a[i] / pre) != 0)
      cnt++;
    else
      pre = a[i];
  return min(2, cnt);
}
int d[N];
int ck() {
  int ss = 0;
  for (int i = 1; i <= n; i++)
    if (a[i]) d[++ss] = a[i];
  double k = (double)d[2] / d[1];
  for (int i = 2; i <= ss; i++)
    if (sgn(k - (double)d[i] / d[i - 1]) != 0) return 0;
  return 1;
}
int Solve(int cas = 0) {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n == 1) return cout << 0 << endl, 0;
  if (n == 2) {
    if (!a[1] && a[2])
      return cout << 1 << endl, 0;
    else
      return cout << 0 << endl, 0;
  }
  int zero = 0;
  for (int i = 1; i <= n; i++) zero += a[i] == 0;
  if (zero) {
    if (zero == n || (a[1] && zero == n - 1))
      cout << 0 << endl;
    else if (zero == n - 1 || (a[1] && zero == n - 2) || (zero == 1 && ck()))
      cout << 1 << endl;
    else
      cout << 2 << endl;
    return 0;
  }
  int ans = INF;
  ans = min(ans, check((double)a[2] / a[1]));
  ans = min(ans, check((double)a[3] / a[1]));
  ans = min(ans, check((double)a[3] / a[2]));
  cout << ans << endl;
  return 0;
}
void Pre() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  Pre();
  while (Solve())
    ;
  return 0;
}
