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;
bool cmp(int a, int b) { return abs(a) < abs(b); }
int F = -100000;
int a[100010];
bool check(int p, double c, int n) {
  bool ans = 0;
  int bf = p;
  for (int i = p + 1; i < n; i++) {
    if (a[i] == F) continue;
    double x = a[bf] * c - a[i];
    if (fabs(a[bf] * c - a[i]) > 1e-4) {
      ans = 1;
      break;
    }
    bf = i;
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  if (a[0] == 0) {
    int count = 0;
    for (int i = 1; i < n; i++) {
      if (a[i] != 0) count++;
    }
    if (count == n - 1) {
      if (n == 2)
        printf("1\n");
      else if (check(1, 1.0 * a[2] / a[1], n))
        printf("2\n");
      else
        printf("1\n");
    } else
      printf("%d\n", min(count, 2));
    return 0;
  }
  int sb = 0;
  double c = 1.0 * a[1] / a[0];
  if (check(0, c, n)) {
    sb = 1;
    int x = a[0];
    a[0] = F;
    if (a[1] == 0 || check(1, 1.0 * a[2] / a[1], n)) {
      a[0] = x;
      x = a[1];
      a[1] = F;
      if (check(0, 1.0 * a[2] / a[0], n)) {
        a[1] = x;
        x = a[2];
        a[2] = F;
        if (check(0, c, n)) {
          a[2] = x;
          for (int i = 1; i < n; i++) {
            if (fabs(a[i - 1] * c - a[i]) > 1e-4) {
              sb = 1;
              a[i] = F;
              if (check(0, c, n)) sb = 2;
              break;
            }
          }
        }
      }
    }
  }
  printf("%d\n", sb);
  return 0;
}
