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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

import static java.lang.Integer.MAX_VALUE;
import static java.lang.Integer.parseInt;
import static java.lang.Math.min;

public class C_Pride {
    static int n,gcd,ar[],count;
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tk;
        n = parseInt(in.readLine());
        ar = new int[n];
        tk = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            ar[i] = parseInt(tk.nextToken());
            gcd = gcd(gcd,ar[i]);
            if(ar[i] == 1)
                count++;
        }
        if(gcd == 1) {
            if(n == 1){
                System.out.println(0);
                return;
            }
            int ans = MAX_VALUE;
            for (int i = 0; i < n; i++) {
                gcd = ar[i];
                for (int j = i + 1; j < n; j++) {
                    gcd = gcd(gcd,ar[j]);
                    if(gcd == 1)
                        ans = min(ans, (j - i + 1) - 2);
                }
            }
            System.out.println(ans + n - count);
        }
        else
            System.out.println(-1);
    }

    static int gcd(int x,int y){
        return (y == 0) ? x : gcd(y, x % y);
    }
}
