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 a[100005];
int n;
void work() {
  if (n == 1) {
    cout << 0 << endl;
    return;
  }
  if (n == 2) {
    if ((a[1] == 0) && (a[2] != 0))
      cout << 1 << endl;
    else
      cout << 0 << endl;
    return;
  }
  int s0 = 0, i;
  for (i = 1; i <= n; i++)
    if (a[i] == 0) s0++;
  if (s0 == n) {
    cout << 0 << endl;
    return;
  }
  if (s0 == n - 1 && a[1] != 0) {
    cout << 0 << endl;
    return;
  }
  if (s0 == n - 1) {
    cout << 1 << endl;
    return;
  }
  if (s0 == n - 2 && a[1] != 0) {
    cout << 1 << endl;
    return;
  }
  int k, s, x;
  if (a[1] != 0 && a[2] != 0) {
    k = 3;
    s = 2;
    x = a[2];
    while (1) {
      while ((k <= n) && (a[k] * a[1] != x * a[2])) k++;
      if (k > n) break;
      x = a[k];
      s++;
      k++;
    }
    if (s == n) {
      cout << 0 << endl;
      return;
    }
    if (s == n - 1) {
      cout << 1 << endl;
      return;
    }
  }
  if (a[1] != 0 && a[3] != 0) {
    k = 4;
    s = 2;
    x = a[3];
    while (1) {
      while ((k <= n) && (a[k] * a[1] != x * a[3])) k++;
      if (k > n) break;
      x = a[k];
      s++;
      k++;
    }
    if (s == n - 1) {
      cout << 1 << endl;
      return;
    }
  }
  if (a[2] != 0 && a[3] != 0) {
    k = 4;
    s = 2;
    x = a[3];
    while (1) {
      while ((k <= n) && (a[k] * a[2] != x * a[3])) k++;
      if (k > n) break;
      x = a[k];
      s++;
      k++;
    }
    if (s == n - 1) {
      cout << 1 << endl;
      return;
    }
  }
  cout << 2 << endl;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  work();
}
