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.*;
import java.io.*;
public class Q892C {
	
	public static void main(String[] args) {
		FastScanner I=new FastScanner();
		Output O=new Output();
		int n=I.nextInt();
		int[] arr=new int[n];
		int gcdd=0, cnt=0;
		for(int i=0;i<n;i++) {
			arr[i]=I.nextInt();
			gcdd=gcd(gcdd,arr[i]);
			if(arr[i]==1) cnt++;
		}
		if(gcdd>1) {
			O.pln(-1);
			return;
		}
		if(cnt>=1) {
			O.pln(n-cnt);
		}
		else {
			int ans=n;
			for(int i=0;i<n;i++) {
				int g=arr[i];
				for(int j=i;j<n;j++){
					g=gcd(g,arr[j]);
					if(g==1) ans=Math.min(ans,j-i);
				}
				
			}
			O.pln(n-1+ans);
		}
		
		
	}
	public static int gcd(int a, int b) {
		while (b > 0) {
			int c = a;
			a = b;
			b = c % b;
		}
		return a;
	}
	
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				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());};
	}

	static class Output{
		PrintWriter w=new PrintWriter(System.out);
		void pln(int x) {w.println(x); w.flush();}
		void pln(long x) {w.println(x); w.flush();}
		void pln(String x) {w.println(x); w.flush();}
		void pln(char x) {w.println(x); w.flush();}
		void pln(StringBuilder x) {w.println(x); w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
	}

}
