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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
 * Created by Tejas on 28-07-2018.
 */
public class Main {

    static int a[];
    static int oneCount=0;
    public static void main(String[]args)throws IOException{
        BufferedReader bufferedReader=new BufferedReader(new InputStreamReader(System.in));
        int N=Integer.parseInt(bufferedReader.readLine());
        int gcd=0;
        String temp[]=bufferedReader.readLine().split(" ");

        a=new int[N];
        for (int i = 0; i < N; i++) {
            a[i]=Integer.parseInt(temp[i]);
            if(a[i]==1)
                oneCount++;
            if(i==0)
                gcd=a[0];
            gcd=gcd(gcd,a[i]);
        }
        if(gcd!=1)
            System.out.println(-1);
        else if(oneCount>0)
            System.out.println(N-oneCount);
        else System.out.println(solve(N));
    }

    private static int solve(int N) {
        int count=0;
        for (int i = 0; i < N-1; i++) {
            int gcd=gcd(a[i],a[i+1]);
            if(gcd==1)
                return N;
        }

        for (int i = 0; i < N-1; i++) {
            int gcd=gcd(a[i],a[i+1]);
            if(gcd==1)
                return count+N;
            if(gcd!=a[i+1]){
                count++;
                a[i+1]=gcd;
            }
        }
        return count;
    }

    private static int gcd(int a,int b){
        if(b==0)
            return a;
        return gcd(b,a%b);
    }
}