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.Scanner;

/**
 * @Created by sbhowmik on 05/02/19
 */
public class Pride {

    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    static public int findMinimumMoves(int arr[]) {

        int ones = 0;
        for (int i = 0; i < arr.length; i++) {
            if(arr[i] == 1) {
                ones++;
            }
        }
        if(ones!=0) {
            return arr.length - ones;
        }

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

    }
    public static void main(String []args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int i = 0;
        int arr[] = new int[n];
        while ( i < n) {
            arr[i++] = scanner.nextInt();
        }
        System.out.println(findMinimumMoves(arr));
    }
}
