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;
void ans(int x) {
  printf("%d\n", x);
  exit(0);
}
bool geom(vector<int> a) {
  if (((int)(a).size()) == 1) return true;
  if (a[0] == 0) {
    for (int i = 0; i < ((int)(a).size()); ++i)
      if (a[i] != 0) return false;
    return true;
  }
  bool mul;
  if (a[1] == 0 || abs(a[1]) >= abs(a[0]))
    mul = true;
  else
    mul = false;
  long long b, q = a[0];
  if (mul)
    b = a[1] / a[0];
  else
    b = a[0] / a[1];
  for (int i = 1; i < ((int)(a).size()); ++i) {
    if (mul)
      q *= b;
    else
      q /= b;
    if (a[i] != q) return false;
  }
  return true;
}
int n;
int a[100005];
int calc(int b, int s) {
  int res = 0;
  int t = a[s];
  for (int i = s + 1; i < n; ++i)
    if (t * b != a[i])
      ++res;
    else
      t *= b;
  return res;
}
void solve_small() {
  vector<int> s;
  for (int i = 0; i < n; ++i) s.push_back(a[i]);
  if (geom(s)) ans(0);
  for (int i = 0; i < n; ++i) {
    vector<int> s;
    for (int j = 0; j < n; ++j)
      if (i != j) s.push_back(a[j]);
    if (geom(s)) ans(1);
  }
}
void solve() {
  if (calc(0, 0) == 0) ans(0);
  if (calc(1, 0) == 0) ans(0);
  if (calc(-1, 0) == 0) ans(0);
  if (calc(0, 0) == 1) ans(1);
  if (calc(1, 0) == 1) ans(1);
  if (calc(-1, 0) == 1) ans(1);
  if (calc(0, 1) == 0) ans(1);
  if (calc(1, 1) == 0) ans(1);
  if (calc(-1, 1) == 0) ans(1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  if (n <= 16) {
    solve_small();
    ans(2);
  }
  solve();
  ans(2);
  return 0;
}
