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.Scanner;

public class P892C
{
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = scan.nextInt();
		int ones = 0;
		for (int i = 0; i < n; i++)
		{
			if (arr[i] == 1)
				ones++;
		}
		if (ones > 0)
		{
			System.out.println(n-ones);
			return;
		}
		
		int min = Integer.MAX_VALUE;
		for (int i = 0; i < n-1; i++)
		{
			//left
			{
				int gcd = arr[i+1];
				int idx = i;
				int ops = 0;
				while (idx >= 0 && gcd != 1)
				{
					gcd = gcd(gcd, arr[idx]);
					ops++;
					idx--;
				}
				if (gcd == 1)
					min = Math.min(min, ops);
			}
			//right
			{
				int gcd = arr[i];
				int idx = i+1;
				int ops = 0;
				while (idx < n && gcd != 1)
				{
					gcd = gcd(gcd, arr[idx]);
					ops++;
					idx++;
				}
				if (gcd == 1)
					min = Math.min(min, ops);
			}
		}
		System.out.println(min == Integer.MAX_VALUE ? -1 : min+n-1);
	}
	static int gcd(int a, int b)
	{
		return b == 0 ? a : gcd(b, a%b);
	}
}
