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;
inline int cmp(double a, double b = 0.0) {
  if (fabs(a - b) <= (1e-9)) return 0;
  if (a < b) return -1;
  return 1;
}
int v[100100];
int n;
int go(double q, int beg) {
  int last = beg - 1;
  int cnt = 0;
  for (int i = beg; i < n; i++) {
    if (cmp(v[last] * q, v[i]) == 0) {
      last = i;
    } else {
      cnt++;
      if (cnt > 1) return 100;
    }
  }
  return cnt;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i];
  if (n == 1)
    cout << 0 << endl;
  else if (n == 2) {
    if (v[0] != 0)
      cout << 0 << endl;
    else if (v[0] == 0 && v[1] == 0)
      cout << 0 << endl;
    else
      cout << 1 << endl;
  } else {
    int res = 2;
    res = min(res, go(0, 1));
    if (v[0] != 0) res = min(res, 0 + go((double)v[1] / (double)v[0], 2));
    if (v[0] != 0) res = min(res, 1 + go((double)v[2] / (double)v[0], 3));
    if (v[1] != 0) res = min(res, 1 + go((double)v[2] / (double)v[1], 3));
    cout << res << endl;
  }
  return 0;
}
