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;
template <class T>
T abs(T x) {
  return x > 0 ? x : (-x);
}
template <class T>
T sqr(T x) {
  return x * x;
}
int isgeom(const vector<long long>& v) {
  if (((int)(v).size()) < 2) return 0;
  long long x = v[1], y = v[0];
  if (y == 0 && x != 0) return 2;
  if (y == 0) {
    int cnt = 0;
    for (int i = 2; i < ((int)(v).size()); ++i)
      if (v[i] != 0) ++cnt;
    if (cnt == 0) return 0;
    if (cnt == 1) return 1;
    return 2;
  }
  int cnt = 0;
  long long px = v[1], py = 1;
  for (int i = 2; i < ((int)(v).size()); ++i) {
    if (cnt >= 2) break;
    long long tmp = px * x;
    if (abs(tmp) % (y * py) != 0) {
      ++cnt;
      continue;
    }
    tmp /= (y * py);
    if (tmp != v[i])
      ++cnt;
    else
      px = v[i], py = 1;
  }
  if (cnt == 0) return 0;
  if (cnt == 1) return 1;
  return 2;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    v[i] = x;
  }
  int t = isgeom(v);
  if (t != 2) {
    cout << t << "\n";
    return 0;
  }
  vector<long long> w;
  for (int i = 1; i < ((int)(v).size()); ++i) w.push_back(v[i]);
  if (isgeom(w) == 0) {
    printf("1\n");
    return 0;
  }
  w.clear();
  for (int i = 0; i < ((int)(v).size()); ++i)
    if (i != 1) w.push_back(v[i]);
  if (isgeom(w) == 0) {
    printf("1\n");
    return 0;
  }
  printf("2\n");
  return 0;
}
