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;
const long long inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double pi = 3.14159265358979;
const double eps = 1e-8;
const int maxn = 100500;
double a[maxn];
int n;
int check(int start, int skip, double b) {
  double cur = a[start];
  bool kick = false;
  for (int i = start + 1; i < n; i++) {
    if (i == skip) continue;
    cur *= b;
    if (abs(a[i] - cur) > eps) {
      if (kick) return 2;
      kick = true;
      if (abs(b) > 0) cur /= b;
    }
  }
  if (kick) return 1;
  return 0;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n <= 2) {
    if (a[0] == 0 && a[1] == 0) {
      puts("0");
      return 0;
    }
    if (a[0] == 0) {
      puts("1");
      return 0;
    }
    puts("0");
    return 0;
  }
  int aa = check(0, -1, a[0] == 0 ? 0 : a[1] / a[0]);
  if (aa < 2) {
    cout << aa << endl;
    return 0;
  }
  aa = check(0, 1, a[0] == 0 ? 0 : a[2] / a[0]);
  if (aa == 0) {
    cout << 1 << endl;
    return 0;
  }
  aa = check(1, 0, a[1] == 0 ? 0 : a[2] / a[1]);
  if (aa == 0) {
    cout << 1 << endl;
    return 0;
  }
  cout << 2 << endl;
  return 0;
}
