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;

public final class PrideCf {
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        int arr[]=new int[n];
        String st[]=br.readLine().split(" ");
        int cnt=0;
        for(int i=0;i<n;i++){
            arr[i]=Integer.parseInt(st[i]);
            if(arr[i]==1) cnt++;
        }
        if(cnt>0){
            System.out.println(n-cnt);
        }else {
            int gcd=arr[0];
            for(int i=1;i<n;i++){
                gcd=Gcd(gcd,arr[i]);
            }
            if(gcd==1){
                int ans=helper(arr,n);
                System.out.println(ans);
            }else
                System.out.println("-1");
        }
    }

    private static int helper(int[] arr, int n) {

        int ans=Integer.MAX_VALUE;
        for(int i=0;i<n-1;i++){
            int gcd=arr[i];
            for(int j=i+1;j<n;j++){
                gcd=Gcd(gcd,arr[j]);
                if(gcd==1){
                    ans=Math.min(ans,j-i);
                    break;
                }
            }
        }
        return ans+n-1;
    }

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