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 A {
	static FastReader sc=null;

	public static void main(String[] args) {
		sc=new FastReader();
		int n=sc.nextInt();
		int a[]=sc.readArray(n);
		int c=0;
		for(int e:a)if(e==1)c++;
		int min=n+1;
		for(int i=0;i<n;i++) {
			int gcd=a[i];
			for(int j=i+1;j<n;j++) {
				gcd=gcd(gcd,a[j]);
				if(gcd==1) {
					min=Math.min(j-i+1, min);
				}
			}
		}
		if(c>0)System.out.println(n-c);
		else if(min==n+1)System.out.println(-1);
		else System.out.println(min+n-2);
		
		
		
			
	}
	
	
 
	
	static int[] reverseSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al,Collections.reverseOrder());
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	static int gcd(int a,int b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static long gcd(long a,long b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static void reverse(int a[]) {
		int n=a.length;
		int b[]=new int[n];
		for(int i=0;i<n;i++)b[i]=a[n-1-i];
		for(int i=0;i<n;i++)a[i]=b[i];
	}
	static int[] ruffleSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al);
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	
	
	
	static void print(int a[]) {
		for(int e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	static void print(long a[]) {
		for(long e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
        int[] readArray(int n) {
    		int a[]=new int [n];
    		for(int i=0;i<n;i++) {
    			a[i]=sc.nextInt();
    		}
    		return a;
    	}
    } 
}
