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>
#pragma comment(linker, "/STACK:836777216")
using namespace std;
int n;
int a[111111];
vector<int> v;
bool good(vector<int> &v, int c, int z = 1) {
  for (int i(1); i < (v.size()); i++)
    if (v[i] * z != v[i - 1] * c) return 0;
  return 1;
}
bool g(vector<int> &v) {
  return (v.size() <= 1 || good(v, 0) || good(v, 1) || good(v, -1) ||
          v[1] && v[0] && good(v, v[1], v[0]));
}
bool tr(vector<int> &v, int c) {
  int pos = 1;
  bool f = 0;
  int now = v[0];
  for (int i(1); i < (v.size()); i++) {
    now *= c;
    while (i < v.size() && v[i] != now) {
      if (f) return 0;
      f = 1;
      i++;
    }
  }
  return 1;
}
bool can(vector<int> &v) { return tr(v, 1) || tr(v, -1) || tr(v, 0); }
int main() {
  cin >> n;
  v.resize(n);
  for (int i(0); i < (n); i++) scanf("%d", a + i), v[i] = a[i];
  if (g(v)) {
    puts("0");
    return 0;
  }
  if (v.size() < 50) {
    for (int i(0); i < (v.size()); i++) {
      vector<int> t = v;
      t.erase(t.begin() + i);
      if (g(t)) {
        puts("1");
        return 0;
      }
    }
    puts("2");
    return 0;
  }
  vector<int> tt = v;
  tt.erase(tt.begin());
  if (can(v) || g(tt)) {
    puts("1");
    return 0;
  }
  puts("2");
  return 0;
}
