You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say x and y, and replace one of them with gcd(x, y), where gcd denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples

Input

5
2 2 3 4 6


Output

5


Input

4
2 4 6 8


Output

-1


Input

3
2 6 9


Output

4

Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

  * [2, 2, 3, 4, 6]. 
  * [2, 1, 3, 4, 6]
  * [2, 1, 3, 1, 6]
  * [2, 1, 1, 1, 6]
  * [1, 1, 1, 1, 6]
  * [1, 1, 1, 1, 1]



We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
import java.util.*;
import java.io.*;
public class a{
    static int gcd(int a,int b){
        if(b==0)return a;
        return gcd(b,a%b);
    }
    public static void main(String [] args){
      Scanner sc=new Scanner(System.in);
      int n=sc.nextInt();
      int [] a=new int[n];
      for(int i=0;i<n;i++){
          a[i]=sc.nextInt();
      }
      int cnt_1=0;
     for(int i=0;i<n;i++){
         if(a[i]==1)cnt_1++;
     }
      if(cnt_1>0){
          System.out.println(n-cnt_1);
          return;
      }

      int min=Integer.MAX_VALUE;
      for(int i=0;i<n;i++){
          int gd=a[i];
          for(int j=i+1;j<n;j++){
              gd=gcd(gd,a[j]);
              if(gd==1){
                  min=Math.min(j-i+(n-1),min);
              }
          }
      }
      if(min==Integer.MAX_VALUE){
          System.out.println(-1);
      }
      else{
          System.out.println(min);
      }
     

     
    }
}