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 MAXN = 100000 + 5;
const double EPS = 1e-3;
int n;
int t[MAXN];
int g(double q) {
  bool ok = true;
  for (int i = 1; i < n; i++)
    if (fabs(double(t[i] - t[i - 1] * q)) > EPS) ok = false;
  if (ok) return 0;
  ok = true;
  for (int i = 2; i < n; i++)
    if (fabs(double(t[i] - t[i - 1] * q)) > EPS) ok = false;
  if (ok) return 1;
  ok = true;
  int prev = t[0];
  for (int i = 1; i < n; i++)
    if (fabs(double(t[i] - prev * q)) > EPS) {
      if (!ok) return 2;
      ok = false;
    } else
      prev = t[i];
  return 1;
}
int f() {
  if (n == 1) return 0;
  double q1 = 0, q2 = 0;
  if (t[0] != 0) q1 = double(t[1]) / t[0];
  if (t[n - 2] != 0) q2 = double(t[n - 1]) / t[n - 2];
  return min(g(q1), g(q2));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &t[i]);
  printf("%d\n", f());
}
