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.math.*;
public class Soln{
	static int gcd(int a,int b){
		while(b>0){
			int c=b;
			b=a%b;
			a=c;
		}
		return a;
	}
	public static void main(String args[])throws Exception{
		int n=nextInt();
		int a[]=new int[n];
		int ones=0;
		for(int i=0;i<n;i++){
			a[i]=nextInt();
			if(a[i]==1)
				ones++;
		}
		if(ones>0){
			System.out.println(n-ones);
			return;
		}
		else{
			int smallest=n+1;
			for(int i=0;i<n;i++){
				int gcd=0;
				for(int j=i;j<n;j++){
					gcd = gcd(a[j],gcd);
					if(gcd==1){
						smallest =Math.min(smallest, j-i);
						break;
					}
				}
			}
			if(smallest==n+1)
				System.out.println(-1);
			else
				System.out.println(smallest+n-1);
		}
	}
	static int nextInt()throws IOException{
		InputStream in=System.in;
        int ans=0;
        boolean flag=true;
        byte b=0;
        while ((b>47 && b<58) || flag){
            if(b>=48 && b<58){
                ans=ans*10+(b-48);
                flag=false;
            }
            b=(byte)in.read();
        }
        return ans;
    }
}