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.lang.*;
public class Pride
{
    public static void main(String args[] ) throws Exception 
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        
        long[] arr=new long[n];
        long x;
        int count=0;
        arr[0]=sc.nextLong();
        x=arr[0];
        if(x==1) 
        {
        	count++;
        }
        for(int j=1;j<n;j++)
        {
            arr[j]=sc.nextLong();
            x=gcd(x,arr[j]);
            if(arr[j]==1)
            {
                count++;
            }
        }
        
        if(x!=1)
        {
            System.out.println(-1);
            
        }
        else{
        if(count>0){
            System.out.println(n-count);
        }
        else
        {
            int min=Integer.MAX_VALUE;
            int len=0;
            for(int i=0;i<n-1;i++)
            {
                long y=arr[i];
                for(int j=i+1;j<n;j++)
                {
                  y=gcd(arr[j],y); 
                 
                   if(y==1)
                   {
                        if(j-i + 1 < min)
                        {
                            min = j-i+1;
                            
                        }
                    }
                }
            }
            
            System.out.println(min + n -2);
        }
        }
    }
    
    static long gcd(long a,long b)
    {
        if(b==0)
        {
            return a;
        }
        
       return gcd(b,a%b);
    }
}
