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 Abc {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		
		int ar[] = new int[n];
		int fOne = 0;
		for(int i = 0; i < n; i++) {
			ar[i] = in.nextInt();
			if(ar[i] == 1) {
				fOne ++;
			}
		}
		if(fOne > 0) {
			System.out.println(n - fOne);
			return;
		}
		int dp[][] = new int[n][n];
		for(int i = 0; i < n;i ++) {
			dp[i][i] = ar[i];
		}

		int minLen = n + 1;
		for(int i = 1; i < n; i ++) {
			for(int j = i - 1; j >= 0; j --) {
				dp[j][i] = gcd(dp[j + 1][i], ar[j]);
				if(i - j < minLen && dp[j][i] == 1) {
					minLen = i - j;
				} 
			}
		}

		if(minLen == n + 1) {
			System.out.println(-1);
		} else {
			System.out.println(minLen + n - 1);
		}
	}
	public static int gcd(int a, int b) {
		if(b == 0) {
			return a;
		} else {
			return gcd(b, a % b);
		}
	}
}