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 c {
	static int gcd(int a, int b)
	{
	  if(a == 0 || b == 0) return a+b; // base case
	  return gcd(b,a%b);
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int numbers[] = new int[n];
		int ones = 0;
		for(int i = 0; i < n; i++) {
			numbers[i] = in.nextInt();
			if(numbers[i] == 1) ones++;
		}
		if(ones > 0) {
			System.out.println(n - ones);
			return ;
		}
		int best = 300000;
		for(int i = 0; i < n; i++) {
			int g = 0;
			for(int j = i; j < n; j++) {
				g = gcd(g, numbers[j]);
				if(g == 1) {
					best =Math.min((j - i) + n - 1, best);
					break;
				}
			}
		}
		if(best == 300000) {
			System.out.println(-1);
		} else {
			System.out.println(best);
		}
	}
}
