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:640000000")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)1e5 + 10;
int n;
vector<int> v;
int a[maxn];
int check(int start) {
  if (v[0] == 0) {
    if (v[1] == 0) {
      int res = 2;
      for (int i = start; i < n; i++)
        if (a[i] == 0) {
          res++;
        }
      return res;
    } else {
      return 0;
    }
  } else {
    if (v[1] == 0) {
      int res = 2;
      for (int i = start; i < n; i++)
        if (a[i] == 0) {
          res++;
        }
      return res;
    } else {
      int y = v[0];
      int x = v[1];
      for (int i = start; i < n; i++) {
        if (v.back() * x == a[i] * y) {
          v.push_back(a[i]);
        }
      }
      return (int)(v).size();
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  if (n == 2) {
    if (a[0] != 0 || a[1] == 0) {
      cout << 0 << endl;
    } else {
      cout << 1 << endl;
    }
    return 0;
  }
  v.push_back(a[0]);
  v.push_back(a[1]);
  int r = check(2);
  if (r == n) {
    cout << 0 << endl;
    return 0;
  }
  if (r == n - 1) {
    cout << 1 << endl;
    return 0;
  }
  v.clear();
  v.push_back(a[0]);
  v.push_back(a[2]);
  r = check(3);
  if (r == n - 1) {
    cout << 1 << endl;
    return 0;
  }
  v.clear();
  v.push_back(a[1]);
  v.push_back(a[2]);
  r = check(3);
  if (r == n - 1) {
    cout << 1 << endl;
    return 0;
  }
  cout << 2 << endl;
  return 0;
}
