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;
vector<int> in;
int n;
bool check_0(int ind) {
  for (int i = ind; i < n; i++)
    if (in[i] != 0) return false;
  return true;
}
bool check() {
  if ((in[0] == 0 && in[1] == 0) || in[1] == 0) {
    if (check_0(1))
      return true;
    else
      return false;
  }
  double d = in[1] * 1. / in[0];
  bool x = true;
  for (int i = 2; i < n; i++)
    if (in[i - 1] == 0 || in[i] * 1. / in[i - 1] != d) {
      x = false;
      break;
    }
  if (x) return true;
  return false;
}
bool check1() {
  if ((in[1] == 0 && in[2] == 0) || in[1] == 0) {
    if (check_0(2))
      return true;
    else
      return false;
  }
  double d = in[2] * 1. / in[1];
  bool x = true;
  for (int i = 3; i < n; i++)
    if (in[i] * 1. / in[i - 1] != d) {
      x = false;
      break;
    }
  if (x) return true;
  return false;
}
bool check2() {
  if ((in[0] == 0 && in[2] == 0) || in[2] == 0) {
    if (check_0(2))
      return true;
    else
      return false;
  }
  double d = in[2] * 1. / in[0];
  bool x = true;
  for (int i = 3; i < n; i++)
    if (in[i] * 1. / in[i - 1] != d) {
      x = false;
      break;
    }
  if (x) return true;
  return false;
}
bool check3() {
  if (in[0] == 0) {
    bool t = true, r = true;
    for (int i = 1; i < n; i++)
      if (in[i] != 0) {
        if (t)
          t = false;
        else
          return false;
      }
    return true;
  }
  if (in[1] == 0) {
    bool t = true, r = true;
    for (int i = 2; i < n; i++)
      if (in[i] != 0) {
        if (t)
          t = false;
        else
          return false;
      }
    return true;
  }
  double d = in[1] * 1. / in[0];
  bool x = true, h = true;
  for (int i = 2; i < n; i++)
    if (in[i] * 1. / in[i - 1] != d) {
      if (h) {
        if (i == n - 1 || (in[i + 1] * 1. / in[i - 1]) == d) {
          i++;
          h = false;
          continue;
        } else {
          x = false;
          break;
        }
      } else {
        x = false;
        break;
      }
    }
  if (x) return true;
  return false;
}
int main() {
  scanf("%i", &n);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  in.resize(n);
  for (int(i) = 0; (i) < n; (i)++) scanf("%i", &in[i]);
  if (n == 2) {
    if (in[0] != 0 || (in[0] == 0 && in[1] == 0))
      printf("0\n");
    else
      printf("1\n");
    return 0;
  }
  if (check()) {
    printf("0\n");
    return 0;
  }
  if (check1() || check2() || check3()) {
    printf("1\n");
    return 0;
  }
  printf("2\n");
  return 0;
}
