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.*;
public class MyClass {
   public static int gcd(int a, int b){
       if(b==0) return a;
       else return gcd(b,a%b);
   }
    public static void main(String args[]) {
     Scanner in = new Scanner(System.in);
      int n=in.nextInt();
      int a[]=new int[n];
      int flag=0;
      for(int i=0;i<n;i++){ a[i]=in.nextInt();  if(a[i]==1) flag++;}
      int g=a[0];
      for(int i=1;i<n;i++) g=gcd(g,a[i]);
      if(g!=1)
      System.out.println("-1");
      else {
          int min=n+1;
          for(int i=0;i<n-1;i++){ int temp=a[i],c=0;
              for(int j=i+1;j<n;j++){
                  temp=gcd(temp,a[j]); c++;
                  if(temp==1) break;
              }if(temp==1) min=Math.min(min,c);
          }if(flag!=0) min=0; else flag=1;
          System.out.println(min+n-flag);
      }
    }
}