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>
int a[100050];
int main(void) {
  double now, k;
  int last;
  int i;
  int n, tot = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) ++tot;
  }
  if (tot == n) {
    printf("0\n");
    return 0;
  }
  if (tot == n - 1 && tot) {
    if (a[1] != 0)
      printf("0\n");
    else
      printf("1\n");
    return 0;
  }
  if (tot == n - 2 && tot) {
    if (a[1])
      printf("1\n");
    else
      printf("2\n");
    return 0;
  }
  if (tot > 1) {
    printf("2\n");
    return 0;
  }
  if (tot == 1) {
    k = -98999999;
    last = -98999999;
    for (i = 1; i <= n; i++) {
      if (a[i] != 0) {
        if (last == -98999999)
          last = a[i];
        else {
          now = (double)a[i] / last;
          if (k == -98999999)
            k = now;
          else if (k != now)
            break;
          last = a[i];
        }
      }
    }
    if (i == n + 1)
      printf("1\n");
    else
      printf("2\n");
    return 0;
  }
  k = -98999999;
  last = -98999999;
  for (i = 1; i <= n; i++) {
    if (last == -98999999)
      last = a[i];
    else {
      now = (double)a[i] / last;
      if (k == -98999999)
        k = now;
      else if (k != now)
        break;
      last = a[i];
    }
  }
  if (i == n + 1) {
    printf("0\n");
    return 0;
  }
  k = -98999999;
  last = -98999999;
  tot = 0;
  for (i = 1; i <= n; i++) {
    if (last == -98999999)
      last = a[i];
    else {
      now = (double)a[i] / last;
      if (k == -98999999) {
        k = now;
        last = a[i];
      } else if (k != now) {
        ++tot;
        if (tot > 1) break;
      } else
        last = a[i];
    }
  }
  if (i == n + 1) {
    printf("1\n");
    return 0;
  }
  k = -98999999;
  last = -98999999;
  for (i = 2; i <= n; i++) {
    if (last == -98999999)
      last = a[i];
    else {
      now = (double)a[i] / last;
      if (k == -98999999)
        k = now;
      else if (k != now)
        break;
      last = a[i];
    }
    if (i == n) {
      printf("1\n");
      return 0;
    }
  }
  printf("2\n");
  return 0;
}
