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


public class C
{
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int[] arr = new int[n];
		int cnt1 = 0;
		for (int i = 0; i < n; i++)
		{
			arr[i] = sc.nextInt();
			if(arr[i] == 1)
				cnt1++;
		}
		int gcd = 0;
		for (int i = 0; i < arr.length; i++)
			gcd = gcd(gcd, arr[i]);
		if(gcd != 1)
			System.out.println(-1);
		else
		{
			if(cnt1 > 0)
				System.out.println(n-cnt1);
			else
			{
				int[][] grange = new int[n][n];
				for (int i = 0; i < n; i++)
				{
					int gtmp = 0;
					for (int j = i; j < n; j++)
					{
						gtmp = gcd(gtmp, arr[j]);
						grange[i][j] = gtmp;
					}
				}
				
				int minSteps = 1000000;
				for (int i = 0; i < n; i++)
				{
					for (int j = i; j < n; j++)
					{
						if(grange[i][j] == 1)
							minSteps = Math.min(minSteps, j-i);
					}
				}
				
				System.out.println(minSteps + n-1);
			}
		}
		
		pw.flush();
		pw.close();
	}
	
	static int gcd(int a, int b)
	{
		if(b == 0)
			return a;
		return gcd(b, a%b);
	}
	static class Scanner
	{
		StringTokenizer st;
		BufferedReader br;
		
		public Scanner()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		public String next() throws IOException
		{
			while(st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		
		public int nextInt() throws IOException
		{
			return Integer.parseInt(next());
		}
		
		public long nextLong() throws IOException
		{
			return Long.parseLong(next());
		}
	}
}
