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 {

    static int gcd(int p,int q)
    {
    int r;
    while(p%q!=0)
    {
    r=q;
    q=p%q;
    p=r;
    }
    return q;
    }
    
    public static void main(String args[]) {
Scanner cro=new Scanner(System.in);
int n,i,q,j,cal,one;
one=0;
n=cro.nextInt();
int g[]=new int[n-1];
j=cro.nextInt();
cal=j;
if(j==1) one++;
for(i=1;i<n;i++)
{
q=j;
j=cro.nextInt();
if(j==1) one++;
g[i-1]=gcd(q,j);
cal=gcd(cal,j);
}
if(cal>1) System.out.println(-1);
else
{
i=1;
j=0;
q=n*2;
if(n>1&&one==0) cal=g[0];
else q=0;
while(i<n-1&&q!=0)
{
cal=gcd(cal,g[i]);
j++;
if(cal==g[i]) j=0;
if(cal==1&&j<q)
{
q=j;
j=0;
}
if(cal==1) cal=g[i];
i++;
}
System.out.println(n-one+q);
}
}       
}