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.*;

public class Main {
    public static int gcd(int a, int b){
        if(b == 0) return a;
        if(b > a) return gcd(b, a);
        return gcd(b, a % b);
    }
    public static void main(String[] arg) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] array = new int[n];
        for(int i = 0; i < n; i++) array[i] = scan.nextInt();

        int oneCount = 0;
        for(int i = 0; i < n; i++) oneCount += array[i] == 1?1:0;

        if(oneCount > 0){
            System.out.println((n - oneCount));
            return;
        }

        int minStepToGenOne = Integer.MAX_VALUE;
        int[][] dp = new int[n][n];
        for(int i = 0; i < n; i++) dp[i][i] = array[i];

        for(int i = 0; i < n; i++){
            for(int j = i + 1; j < n; j++){
                dp[i][j] = gcd(dp[i][j - 1], array[j]);
                if(dp[i][j] == 1)
                    minStepToGenOne = Math.min(minStepToGenOne, j - i);
            }
        }

        if(minStepToGenOne == Integer.MAX_VALUE){
            System.out.println(-1);
        }else{
            System.out.println((n - 1 + minStepToGenOne));
        }
    }
}