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 A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            int n=Integer.parseInt(br.readLine());
            String s1[]=br.readLine().split(" ");
            int a[]=new int[n];
            int cnt=0;
            for(int i=0;i<n;i++)
            {
             a[i]=Integer.parseInt(s1[i]);
             if(a[i]==1)
              cnt++;
            }
            
            if(cnt>0)
             System.out.println(n-cnt);
            else
            {
                  int dp[][]=new int[n][n];
                  int min=Integer.MAX_VALUE;
                  for(int i=0;i<n;i++)
                  {
                        dp[i][i]=a[i];
                        for(int j=i+1;j<n;j++)
                        {
                              dp[i][j]=gcd(dp[i][j-1],a[j]);
                              if(dp[i][j]==1)
                               min=Math.min(min,j-i);
                        }
                  }
                  
                  if(min==Integer.MAX_VALUE)
                   System.out.println("-1");
                  else
                   System.out.println(n-1+min);
            }
      }
      
      public static int gcd(int a,int b)
      {
            if(b==0)
             return a;
            return gcd(b,a%b);
      }
}