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;
vector<int> a;
vector<int> b;
int n, kol0, x;
bool f(vector<int> a) {
  for (int i = 2; i < a.size(); ++i)
    if (a[i] * a[0] != a[i - 1] * a[1]) return 0;
  return 1;
}
int kol(vector<int> a) {
  int k = 0;
  for (int i = 0; i < a.size(); ++i) k += (a[i] == 0 ? 1 : 0);
  return k;
}
void without(int k) {
  vector<int> b;
  for (int i = 0; i < a.size(); ++i)
    if (i != k) b.push_back(a[i]);
  if (f(b)) {
    cout << 1;
    exit(0);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x;
    a.push_back(x);
  }
  if (a[0] != 0 && a[1] == 0) a.erase(a.begin());
  n = a.size();
  if (a[0] == 0) {
    kol0 = kol(a);
    if (kol0 == n) {
      cout << 0;
      return 0;
    }
    if (kol0 == n - 1) {
      cout << 1;
      return 0;
    }
    if (kol0 <= n - 2 && kol0 >= 2) {
      cout << 2;
      return 0;
    }
  }
  if (a[0] == 0) without(0);
  if (f(a)) {
    cout << 0;
    return 0;
  }
  without(0);
  without(1);
  for (int i = 2; i < n; ++i)
    if (a[i] * a[0] != a[i - 1] * a[1]) {
      without(i);
      without(i - 1);
      if (i + 1 < n) without(i + 1);
    }
  cout << 2;
  return 0;
}
