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, u[100005];
double a[100005];
int ch() {
  if (n == 1) return 1;
  if (n == 2 && a[0] != 0) return 1;
  if (a[0] == 0) {
    for (int i = 0; i < n; i++)
      if (a[i] != 0) return 0;
    return 1;
  }
  double b = a[1] / a[0];
  double now = a[1];
  for (int i = 2; i < n; i++) {
    now *= b;
    if (a[i] != now) return 0;
  }
  return 1;
}
int ch2(int st, double c, double cb, int del) {
  if (n == 2) return 1;
  if (n == 3 && c != 0) return 1;
  if (c == 0) {
    if (cb != 0) return 0;
    for (int i = 3; i < n; i++)
      if (a[i] != 0) {
        if (del) return 0;
        del = 1;
      }
    return 1;
  }
  double b = cb / c;
  double pre = cb;
  for (int i = st; i < n; i++) {
    if (a[i] != pre * b) {
      if (del) return 0;
      del = 1;
    } else
      pre *= b;
  }
  return 1;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%lf", &a[i]);
      u[i] = 1;
    }
    if (ch())
      puts("0");
    else if (ch2(3, a[1], a[2], 1) || ch2(3, a[0], a[2], 1) ||
             ch2(2, a[0], a[1], 0))
      puts("1");
    else
      puts("2");
  }
}
