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.math.*;
import java.util.*;
import javax.swing.Spring;

public class Main {
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	public static void main(String[] args) {
		final int maxn = (int) (1e6 + 5);
//		int[] vis = new int[maxn];
//		int[] book = new int[maxn];
		int[] a = new int[maxn];
//		int[] b = new int[maxn];
		Scanner cin = new Scanner(System.in);
//		Map map = new HashMap();
//		Set<Character> s = new HashSet<>();
//		Set<Set<Character>> ss = new HashSet<>();
		// System.out.println(String2);
		int n = cin.nextInt();
//		String String2 = cin.nextLine();
//		char []String1 = String2.toCharArray();//字符串拆分成数组
		int cnt = 0;
		for (int i = 1; i <= n; i++) {
			a[i] = cin.nextInt();
			if (a[i] == 1)
				cnt++;
		}
		int ans = 0;
		int minn = maxn;
		for (int i = 1; i <= n - 1; i++) {
			int t = a[i];
			for (int j = i + 1; j <= n; j++) {
				t = gcd(t, a[j]);
				ans++;
				if (t == 1)
					{
					minn = Math.min(minn, ans);
					ans = 0;
					break;
					}
			}
			ans = 0;
		}
		//System.out.println(minn);
		if (cnt != 0)
			System.out.println(n - cnt);
		else if (minn  == maxn)
			System.out.println(-1);
		else
			System.out.println(n + minn - 1);
	}
}
		 	      				  	 		 	 	 	 			