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.Arrays;
import java.util.Scanner;

public class Pride446C {

	public static void main(String[] args) {
		Scanner s =new Scanner(System.in);
		int n = s.nextInt();
		int arr[] = new int[n];
		int f[] = new int[32000];
		int one = 0;
		for(int i =0;i<n;i++) {
			arr[i] = s.nextInt();
			if(arr[i] == 1) {
				one++; 			
				}
		}
		if(n == 1) {
			if(one>0) {
				System.out.println(0);
			}else {
				System.out.println(-1);
			}
			return;
		}
		
		if(one > 0) {
			System.out.println(n - one);
			return;
		}
		long g = gcd(arr[0],arr[1]);
		
		for(int i = 2;i<n;i++) {
			g = gcd(g,arr[i]);
		}
		
		if(g != 1) {
			System.out.println(-1);
			return;
		}
		
		long ans = Integer.MAX_VALUE;
		for(int l = 0;l<n - 1;l++) {
			long gc = arr[l];
			for(int r = l + 1; r < n;r++) {
				gc = gcd(gc, arr[r]);
				if(gc == 1) {
					ans = Math.min(ans, r - l + n - 1);
					break;
				}
			}
		}
		System.out.println(ans);
		
		


	}
	
	private static long gcd(long a, long b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}


}
