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.*;

public class Pride {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] arr = new int[n];
		int countOne = 0;
		for(int i = 0; i < n; i++){
			arr[i] = scan.nextInt();
			if(arr[i] == 1) countOne++;
		}
		int ans = 0;
		if(countOne == n){
			System.out.println(0);
			return;
		}
		else if(countOne == 0){
			int[][] gcdOfRange = new int[n][n]; //from 0 to n-1
			int len = Integer.MAX_VALUE;
			int left = -1;
			int right = -1;
			for(int i = 0; i < n; i++){
				for(int j = i; j < n; j++){
					if(j == i) gcdOfRange[i][i] = arr[i];
					else gcdOfRange[i][j] = gcd(gcdOfRange[i][j-1], arr[j]);
					//System.out.println(gcdOfRange[i][j]);
					if(gcdOfRange[i][j] == 1 && j-i+1 < len){ //this is the interval
						len = j-i+1;
						left = i;
						right = j;
					}
				}
			}
			if(left == -1 && right == -1){
				System.out.println(-1);
				return;
			}
			ans += len-1;
			arr[left] = 1;
			countOne++;
		}
		System.out.println(ans+n-countOne);
	}

	static int gcd(int a, int b){
		if(b == 0) return a;
		return gcd(b, a%b);
	}
}
