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 CF892C {
	static final int MAX = (int)(Math.sqrt(1e9))+10;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] a = new int[n];
		for(int i = 0 ; i < n ; i++)
			a[i] = scan.nextInt();
		boolean[] isPrime = new boolean[MAX];
		Arrays.fill(isPrime, true);
		isPrime[0] = isPrime[1] = false;
		boolean commonPrime = false;
		for(int i = 2 ; i < MAX ; i++) {
			if(isPrime[i]) {
				commonPrime = true;
				for(int j = 0 ; j < n ; j++)
					if(a[j]%i != 0)
						commonPrime = false;
				for(int j = i+i ; j < MAX ; j+=i)
					isPrime[j] = false;
			}
			// if there is a common prime factor amongst all a[i], then it is impossible
			if(commonPrime) {
				System.out.println(-1);
				return;
			}
		}
		int ones = 0;
		for(int i = 0 ; i < n ; i++)
			if(a[i] == 1)
				ones++;
		if(ones > 0) {
			System.out.println(n-ones);
		} else {
			int res = Integer.MAX_VALUE;
			for(int i = 0 ; i < n ; i++) {
				for(int j = i+1 ; j < n ; j++) {
					if(gcd(a[i], a[j]) == 1) {
						res = Math.min(res, n+(j-i)-1);
					}
				}
			}
			if(res == Integer.MAX_VALUE)
				res = n+1;
			System.out.println(res);
		}
	}
	static int gcd(int a, int b) {
		if(b == 0)
			return a;
		return gcd(b, a%b);
	}
}
