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 Main {
     static int gcd(int a, int b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    } 
    static int findGCD(int arr[], int n) 
    { 
        int result = 0; 
        for (int element: arr){ 
            result = gcd(result, element); 
  
            if(result == 1) 
            { 
               return 1; 
            } 
        } 
  
        return result; 
    } 
  
	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 count=0;
	        for(int i=0;i<n;i++) {
	            if(a[i]==1)
	            count++;
	        }
	        if(count>0) {
	            System.out.println(n-count);
	        }
	        else {
	            int gc=findGCD(a,n);
	            if(gc!=1) 
	            System.out.println("-1");
	            else {
	                int g=0;
	                int res=Integer.MAX_VALUE;
	                for(int i=0;i<n-1;i++) {
	                    g=a[i];
	                    for(int j=i+1;j<n;j++) {
	                        g=gcd(g,a[j]);
	                        if(g==1) {
	                            res=Math.min(res,j-i);
	                            break;
	                        }
	                    }
	                }
	                System.out.println(res+n-1);
	            }
	        }
	    }
           
	} 

