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.*;
import java.io.*;
public class Codechef
{
    static int GCD(int a,int b){
        if(b==0)
        return a;
        return GCD(b,a%b);
    }
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc=new Scanner(System.in);
	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    int t=1;
	    //t=sc.nextInt();
	  //  int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	        int n=sc.nextInt();
	         int a[]=new int[n];
	         int one=0;
	        for(int i=0;i<n;i++){
	            a[i]=sc.nextInt();
	            if(a[i]==1)
	            one++;
	            
	        }
	        int k=0,min=1000000007;;
	        for(int i=0;i<n;i++){
	            int gcd=a[i];
	            for(int j=i;j<n;j++){
	              gcd=GCD(gcd,a[j]);    
	                if(gcd==1){
	                    min=Math.min(j-i,min);
	                    k=1;
	                }
	              //  System.out.println(min);
	            }
	            
	        }
	        
	        if(k==0)
	        System.out.println(-1);
	        else if(one>0){
	            System.out.println(n-one);
	        }
	        else{
	            System.out.println(n-1+min);
	        }
	        
	    }
	    
	}
}
