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;
double org[100010];
int change0() {
  int count = 0;
  double b = org[1] / org[0];
  int bf = 1;
  if (b == 0) {
    for (int i = 2; i < n; i++) {
      if (org[i] != 0) count++;
    }
    return count;
  }
  for (int i = 2; i < n; i++) {
    if (!(fabs(b - org[i] / org[bf]) < 1e-6))
      count++;
    else
      bf = i;
  }
  return count;
}
int change1() {
  if (org[1] == 0) {
    for (int i = 2; i < n; i++)
      if (org[i] != 0) return -1;
    return 1;
  }
  double b = org[2] / org[1];
  for (int i = 3; i < n; i++) {
    if (!(fabs(b - org[i] / org[i - 1]) < 1e-6)) return -1;
  }
  return 1;
}
int change2() {
  double b = org[2] / org[0];
  for (int i = 3; i < n; i++) {
    if (!(fabs(b - org[i] / org[i - 1]) < 1e-6)) return -1;
  }
  return 1;
}
int main(int argc, char *argv[]) {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) scanf("%lf", &org[i]);
    if (n == 1)
      printf("0");
    else {
      if (org[0] == 0) {
        int tmp = change1();
        if (n > 2) {
          int cc = 0;
          for (int i = 1; i < n; i++)
            if (org[i] != 0) cc++;
          if (cc > 1) {
            if (tmp == 1)
              printf("1\n");
            else
              printf("2\n");
          } else if (cc == 1)
            printf("1\n");
          else
            printf("0\n");
        } else if (n == 2) {
          if (org[0] == 0 && org[1] == 0)
            printf("0\n");
          else if (org[0] == 0 || org[1] == 0)
            printf("1\n");
          else
            printf("0\n");
        }
      } else {
        if (n <= 2)
          printf("0\n");
        else {
          int tmp = change0();
          if (tmp >= 0 && tmp <= 1)
            printf("%d\n", tmp);
          else {
            tmp = change1();
            if (tmp == 1)
              printf("%d\n", tmp);
            else {
              tmp = change2();
              if (tmp == 1)
                printf("1\n");
              else
                printf("2\n");
            }
          }
        }
      }
    }
  }
  return 0;
}
