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  Pride
{
	public static void main(String[] args) throws IOException 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out);

		int t=1;

		while(t-->0)
		{

			int n=Integer.parseInt(br.readLine());
			String str[]=br.readLine().trim().split(" ");
			long a[]=new long[n];
			for(int i=0;i<n;i++)
			{
				a[i]=Long.parseLong(str[i]);
			}

			long gcd=0;
			long count=0;
			for(int i=0;i<n;i++)
			{
				if(a[i]==1)
					count++;
				gcd=gcd(gcd,a[i]);
			}


			if(gcd!=1)
				pw.println(-1);
			else if(count>0)
			{
				pw.println(n-count);
			}
			else
			{
				int ans=Integer.MAX_VALUE;
				
				for(int i=0;i<n;i++)
				{
					gcd=0;
					for(int j=i;j<n;j++)
					{
						gcd=gcd(gcd,a[j]);
						if(gcd==1)
						{
							ans=Math.min(ans,Math.abs(i-j));
							break;
						}
					}
				}


				pw.println((n-1)+ans);

			}
		}
		pw.flush();
	}
	static long gcd(long a, long b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }
	
}