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.ArrayList;
import java.util.Scanner;

public class Main {
	static final int maxn = (int)1e9+10;
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int n;
		n = cin.nextInt();
		int[] a = new int[2020];
		int cnt =0;
		for(int i=0;i<n;i++) {
			a[i] = cin.nextInt();
			if(a[i]==1)
				cnt++;
		}
		int ans=maxn;
		if(cnt>0) {
			System.out.println(n-cnt);
		}
		else {
			for(int i=0;i<n;i++) {
				int m=a[i];
				for(int j=i+1;j<n;j++) {
					m=gcd(m,a[j]);
					if(m==1) {
						ans=Math.min(ans,j-i-1 );
						break;
					}
				}
			}
			if(ans==maxn)
				System.out.println(-1);
			else
				System.out.println(n+ans);
		}
	}
	static int gcd(int a,int b) {
		if(b==0)
			return a;
		return gcd(b,a%b);
	}
	
}
