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;

public class Main {
	
	private static int gcd(int a , int b) {
		while (b > 0) {
			int t = a;
			a = b;
			b = t % a;
		}
		return a;
	}
	
	private static int[] input = new int[10000];

	public static void main(String[] args) throws Exception {		
		
		Scanner scan = new Scanner(System.in);
		
		int i , j , n = scan.nextInt() , cnt1 = 0;
		for (i = 0;i < n;i ++) {
			input[i] = scan.nextInt();
			if (input[i] == 1) {
				cnt1 ++;
			}
		}
		if (cnt1 > 0) {
			System.out.println(n - cnt1);
		} else {
			int ans = - 1;
			for (i = 0;i < n;i ++) {
				int value = input[i];
				for (j = i;j < n;j ++) {
					value = gcd(value , input[j]);
					if (value == 1) {			
						int temp = j - i + n - 1;
						if (temp < ans || ans < 0) {
							ans = temp;
						}
					}
				}
			}
			System.out.println(ans);
		}
		
	}
	
	
}








