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.*;
import java.text.*;
import java.util.*;

public class A {
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static Scanner in = new Scanner(new InputStreamReader(System.in));

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

    public static void main(String[] args) {
        while (in.hasNextInt()) {
            int n = in.nextInt();
            int[] sum = new int[n + 5];
            int ans = 0;
            for (int i = 0; i < n; i++) {
                sum[i] = in.nextInt();
                if (sum[i] == 1) {
                    ans++;
                }
            }
            if (ans != 0) {
                out.println(n - ans);
            } else {
                boolean vis = false;
                for (int i = 1; i < n; i++) {
                    for (int j = 0; j < n - i; j++) {
                        sum[j] = gcd(sum[j], sum[j + 1]);
                        if (sum[j] == 1) {
                            vis = true;
                            ans = i + n - 1;
                            break;
                        }
                    }
                    if (vis) {
                        break;
                    }
                }
                out.println(vis ? ans : -1);
            }
            out.flush();
        }
    }
}