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;
int n;
vector<int> v;
bool check(const vector<int> &v, bool flag) {
  if ((int)v.size() == 1) {
    return true;
  }
  if (v[0] == 0) {
    for (int i = 1; i < (int)v.size(); i++) {
      if (v[i]) {
        if (!flag) {
          return false;
        }
        flag = false;
      }
    }
    return true;
  }
  if ((int)v.size() == 2) {
    return true;
  }
  double r = 1.0 * v[1] / v[0];
  double now = v[1];
  for (int i = 2; i < (int)v.size(); i++) {
    if (fabs(now * r - v[i]) >= 1e-9) {
      if (!flag) {
        return false;
      }
      flag = false;
    } else {
      now *= r;
    }
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  if (check(v, 0)) {
    puts("0");
  } else {
    vector<int> t;
    for (int i = 1; i < (int)v.size(); i++) {
      t.push_back(v[i]);
    }
    if (check(t, 0)) {
      puts("1");
    } else {
      t.clear();
      t.push_back(v[0]);
      for (int i = 2; i < (int)v.size(); i++) {
        t.push_back(v[i]);
      }
      if (check(t, 0)) {
        puts("1");
      } else {
        puts(check(v, 1) ? "1" : "2");
      }
    }
  }
}
