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;
const double error = 1e-9;
const int MAXN = 100005;
int a[MAXN];
int c[3][2], cn;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1)
    printf("0\n");
  else if (n == 2) {
    if (a[0] == 0 && a[1] != 0)
      printf("1\n");
    else
      printf("0\n");
  } else {
    int ans = 2;
    cn = 3;
    c[0][0] = a[0], c[0][1] = a[1];
    c[1][0] = a[0], c[1][1] = a[2];
    c[2][0] = a[1], c[2][1] = a[2];
    for (int k = 0; k < cn; k++) {
      int cnt = 0;
      double x = c[k][0], y;
      if (c[k][0] == 0)
        y = 0;
      else
        y = (double)c[k][1] / c[k][0];
      for (int i = 0; i < n; i++)
        if (fabs(x - a[i]) > error)
          cnt++;
        else
          x *= y;
      ans = min(ans, cnt);
    }
    printf("%d\n", ans);
  }
  return 0;
}
