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;
import java.util.ArrayList;

public class C {

    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    ArrayList<Integer> a;
    int n;

    public void run() throws IOException{
        readInput();
        int x, steps;
        int best = 2*n+1;

        int ones = 0;
        for(int i = 0; i<n; i++){
            if(a.get(i) == 1)ones++;
        }
        if(ones>0){
            System.out.println(n-ones);
            return;
        }

        for(int i = 0; i<n; i++){
            x = a.get(i);
            steps = 0;
            for(int j = i+1; j<n; j++){
                if(x == 1){
                    best = Math.min(best, steps + n - 1);
                    break;
                }
                x = gcd(x, a.get(j));
                steps++;
                if(x == 1){
                    best = Math.min(best, steps + n - 1);
                    break;
                }
            }
        }

        if(best == 2*n + 1){
            System.out.println(-1);
        }else {
            System.out.println(best);
        }
    }

    public void readInput() throws IOException{
        n = pint(br.readLine());
        a = new ArrayList<>();
        String[] s= br.readLine().split("\\s+");
        for(int i = 0; i<n; i++){
            a.add(pint(s[i]));
        }
    }

    static int gcd(int a, int b) {
        while (b != 0) {
            int t = a;
            a = b;
            b = t % b;
        }
        return a;
    }

    public static void main(String[] args) throws IOException{
        new C().run();
    }

    public static int pint(String input){
        return Integer.parseInt(input);
    }

    public static long plong(String input){
        return Long.parseLong(input);
    }

    public static double pdouble(String input){
        return Double.parseDouble(input);
    }

}
