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.*;
import java.util.*;
public class Solution {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	//Main
	static long gcd(long a,long b) {
		return b==0?a:gcd(b,a%b);
	}
	public static void main(String args[]) {
		int test=1;
		//test=sc.nextInt();
		while(test-->0) {
			int n=sc.nextInt(),min=Integer.MAX_VALUE,o=0;
			long a[]=new long[n];
			for(int i=0;i<n;i++) {
				a[i]=sc.nextLong();
				if(a[i]==1) o++;
			}
			if(o>0) {
				out.println(n-o);
				continue;
			}
			for(int i=0;i<n;i++) {
				long g=0;
				int j=i;
				for(;j<n;j++) {
					g=gcd(g,a[j]);
					if(g==1) break;
				}
				if(g==1) min=Math.min(min, j-i);
			}
			if(min==Integer.MAX_VALUE) out.println(-1);
			else out.println(min+n-1);
		}
		out.flush();
		out.close();
	}
}
