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
import java.io.*;
import java.util.*;

public class D {

   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);

      int n = sc.nextInt();
      double[] vals = new double[n];
      double[] vals2 = new double[n];
      int numZeros = 0;
      for (int i=0; i<n; i++) {
         vals[i] = vals2[i] = sc.nextInt();
         if (vals[i] == 0) numZeros++;
      }

      if (n == 1 || numZeros == n) {
         System.out.println(0);
         return;
      }

      if (n == 2) {
         System.out.println(vals[0] == 0 ? 1 : 0);
         return;
      }

      if (numZeros == n - 1) {
         System.out.println(vals[0] != 0 ? 0 : 1);
         return;
      }

      if (numZeros == n - 2 && vals[0] != 0) {
         System.out.println(1);
         return;
      }

      if (numZeros > 1) {
         System.out.println(2);
         return;
      }

      boolean deleted = false;
      double rat;
      if (vals[0] == 0) {
         deleted = true;
         rat = vals[2] / vals[1];
      } else {
         rat = vals[1] / vals[0];
      }

      //System.out.format("rat is %f\n", rat);

      final double eps = 1e-8;
      for (int i=1; i<n-1; i++) {
         double r2 = vals[i+1] / vals[i];
         if (r2 != rat) {
            if (deleted) {
               // try again with deleting first
               boolean yay = true;
               rat = vals2[2] / vals2[1];
               for (i=1; i<n-1; i++) {
                  r2 = vals2[i+1] / vals2[i];
                  if (r2 != rat) {
                     yay = false;
                     break;
                  }
               }
               if (yay) {
                  System.out.println(1); // possible
                  return;
               }

               // try again with deleting second
               rat = vals2[2] / vals2[0];
               for (i=2; i<n-1; i++) {
                  r2 = vals2[i+1] / vals2[i];
                  if (r2 != rat) {
                     System.out.println(2); // not possible
                     return;
                  }
               }

               System.out.println(1); // possible
               return;
            } else {
               deleted = true;
               vals[i+1] = vals[i];
            }
         }
      }

      System.out.println(deleted ? 1 : 0);
   }
}
