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 p891A {
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer st;
		int size = Integer.parseInt(f.readLine());
		int[] a = new int[size];
		st = new StringTokenizer(f.readLine());
		for(int i = 0; i < size; i++)
			a[i] = Integer.parseInt(st.nextToken());
		int count = 0;
		int gcd = a[0];
		for(int num : a)
		{
			gcd = gcd(gcd, num);
			if(num == 1)
				count++;
		}
		if(count > 0)
		{
			out.println(size - count);
			out.close();
			return;
		}
		else if(gcd != 1)
		{
			out.println(-1);
			out.close();
			return;
		}
		out.println(minRange(a) + size - 1);
		out.close();
	}
	
	public static int minRange(int[] a)
	{
		int min = a.length;
		for(int r = 0; r < a.length; r++)
		{
			int gcd = a[r];
			for(int c = r + 1; c < a.length; c++)
			{
				gcd = gcd(gcd, a[c]);
				if(gcd == 1)
				{
					min = Math.min(min, c - r);
					break;
				}
			}
		}
		return min;
	}
	
	public static int gcd(int x, int y)
	{
		while(y > 0)
		{
			int temp = y;
			y = x % y;
			x = temp;
		}
		return x;
	}
}