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 Solution {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		System.out.println(minimumMoves(arr, n));
	}

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

	static int minimumMoves(int arr[], int n) {
		int one = 0;
		for (int i = 0; i < n; i++)
			if (arr[i] == 1)
				one++;

		if (one != 0)
			return n - one;
		int minimum = Integer.MAX_VALUE;
		for (int i = 0; i < n; i++) {
			int g = arr[i];
			for (int j = i + 1; j < n; j++) {
				g = gcd(arr[j], g);
				if (g == 1) {
					minimum = Math.min(minimum, j - i);
					break;
				}
			}
		}

		if (minimum == Integer.MAX_VALUE)
			return -1;
		else
			return n + minimum - 1;
	}

}