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.PrintWriter;
import java.util.Scanner;

public class A891 {

    public static void main(String[] args) {
        try (Scanner reader = new Scanner(System.in); PrintWriter writer = new PrintWriter(System.out)) {
            final int n = reader.nextInt();
            final int[] a = new int[n];
            for (int i = 0; i < a.length; i++) {
                a[i] = reader.nextInt();
            }
            int gcd = a[0];
            for (int i = 1; i < a.length; ++i) {
                gcd = gcd(gcd, a[i]);
            }
            if (gcd != 1) {
                writer.println(-1);
            } else if (n == 1) {
                writer.println(0);
            } else {
                int cnt1 = 0;
                for (int i = 0; i < n; ++i) {
                    if (a[i] == 1) cnt1 += 1;
                }
                if (cnt1 > 0) {
                    writer.println(n - cnt1);
                } else {
                    int min = n;
                    for (int i = 0; i < n; ++i) {
                        int g = a[i];
                        for (int j = i + 1; j < n; ++j) {
                            g = gcd(g, a[j]);
                            if (g == 1) {
                                min = Math.min(min, j - i);
                                break;
                            }
                        }
                    }
                    writer.println(min + n - 1);
                }

            }
        }
    }

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

}
