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 Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] temp=br.readLine().trim().split(" ");
		
		int n=Integer.parseInt(temp[0]);
		int[] arr=new int[n];
		temp=br.readLine().trim().split(" ");
		for(int i=0;i<n;i++){
		    arr[i]=Integer.parseInt(temp[i]);
		}
		
		out.println(minOperations(arr));
		
		out.flush();
		out.close();
	}
	public static int minOperations(int[] arr){
	    int n=arr.length;
	    int gcd=arr[0];
	    for(int i=1;i<n;i++){
	        gcd=gcd(gcd,arr[i]);
	    }
	    if(gcd>1){
	        return -1;
	    }
	    int num1s=0;
	    for(int i=0;i<n;i++){
	        if(arr[i]==1){
	            num1s++;
	        }
	    }
	    int ans=Integer.MAX_VALUE;
	    for(int start=0;start<n;start++){
	        gcd=arr[start];
	        for(int end=start;end<n;end++)
	        {
	            gcd=gcd(gcd,arr[end]);
	            if(gcd==1){
	                int size=end-start+1;
	                int operations=size-1;
	                operations+=(n-1-num1s);
	                if(arr[end]==1){
	                    operations++;
	                }
	                ans=Math.min(ans,operations);
	                break;
	            }
	        }
	    }
	    return ans;
	}
	public static int gcd(int a,int b){
	    if(a==0){
	        return b;
	    }
	    if(b==0){
	        return a;
	    }
	    return gcd(b%a,a);
	}
}