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.*;
public class Solution
{
    static int dp[][];
    static int N;
    static int A[];
    public static void main(String ag[])
    {
        Scanner sc=new Scanner(System.in);
        int i,j,k;
        N=sc.nextInt();
        A=new int[N];
        
        for(i=0;i<N;i++)
        A[i]=sc.nextInt();
        
        dp=new int[N+1][N];
        for(i=0;i<=N;i++)
        Arrays.fill(dp[i],-1);
        
        int one=0;
        for(i=0;i<N;i++)
        {
            if(A[i]==1)
            one++;
        }
        if(one>0)
        {
            System.out.println(N-one);
            return;
        }
        for(int len=2;len<=N;len++)
        {
            for(j=0;j+len-1<N;j++)
            {
                if(find(j,len)==1)
                {
                    System.out.println(len-2+N);
                    return;
                }
            }
        }
        System.out.println(-1);
    }
    public static int find(int id,int len)
    {
        if(len==1)
        return A[id];
        
        if(len==2)
        return GCD(A[id],A[id+len-1]);
        
        if(dp[len][id]!=-1)
        return dp[len][id];
        
        return dp[len][id]=GCD(GCD(A[id],A[id+len-1]),find(id+1,len-2));
    }
    
    public static int GCD(int a,int b)
    {
        if(b==0)
        return a;
        return GCD(b,a%b);
    }
    
}