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 Pride {

	static int gcd(int a, int b)
	{
		if(a==0)
			return b;
		return gcd(b%a,a);
	}
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int gcd=0,one=0;
		HashSet<Integer> set=new HashSet<>();
		int a[]=new int[n];
		for(int i=0;i<n;i++)
		{
			a[i]=sc.nextInt();
			if(a[i]==1)
				one++;
			set.add(a[i]);
			gcd=gcd(gcd,a[i]);
		}
		if(gcd!=1)
		{
			System.out.println(-1);
			System.exit(0);
		}
		int min=Integer.MAX_VALUE;
		//Minimum number of steps to convert any one number to 1 = (size of minimum subarray with gcd=1)-1
		//Thus to find this minimum steps we iterate through all possible subarrays since n=2000 at max.
		for(int i=0;i<n;i++)
		{
			int x=0;
			for(int j=i;j<n;j++)
			{
				x=gcd(x,a[j]);
				if(x==1 && (j-i+1)<min)
					min=j-i+1;
			}
		}
		if(one==0)
			one++;
		System.out.println((n-1+min-1-(one-1)));

	}

}
