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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Round446Div1A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter(new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int n=s.nextInt();
		int[][] dp=new int[n][n];
		for(int i=0;i<n;i++) {
			dp[0][i]=s.nextInt();
		}
		int ans =Integer.MAX_VALUE;
		int count=0;
		for(int i=0;i<n;i++) {
			if(dp[0][i]==1) {
				ans=0;
				count++;
			}
		}
		outerloop:for(int i=1;i<n;i++) {
			for(int j=0;j<n-i;j++) {
				dp[i][j]=(int)gcd(dp[i-1][j+1],dp[i-1][j]);
				if(dp[i][j]==1) {
					ans= Integer.min(ans, i);
					break outerloop;
				}
			}
		}
		if(count==0)count++;
		if(ans==Integer.MAX_VALUE)out.println(-1);
		else {
			out.println(ans+n-count);
		}
		out.close();
	}
	
	public static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			//InputStream reads the data and decodes in character stream
			//It acts as bridge between byte stream and character stream
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	
}
