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

public class Temp{  
    static long gcd(long a,long b){
        if(b==0){
            return a;
        }
        else{
            return gcd(b,a%b);
        }
    }
    static PrintWriter w=new PrintWriter(System.out);
    
    public static void main (String[] args)throws IOException{
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        long arr[]=new long[n];
        long gcd[]=new long[n];
        long min=Long.MAX_VALUE;
        int o=0;
        for(int i=0;i<n;i++){
            arr[i]=sc.nextInt();
            min=Math.min(arr[i],min);
            if(arr[i]==1)o++;
        }
        if(o>0)w.println(n-o);
        else{
            int c=100000;
            for(int i=0;i<n;i++){
                long g=arr[i];
                for(int j=i+1;j<n;j++){
                    g=gcd(g,arr[j]);
                    if(g==1)c=Math.min(c,j-i);
                }
            }
            w.println(c==100000?-1:(c+n-1));
            
        }
        
        
        
        
        w.close();
    }
}