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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] arr = new int[n];
            int ones = 0;
            int ans = 0;
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
                if (arr[i] == 1) {
                    ones++;
                }
            }
            if (ones > 0) {
                ans = n - ones;
            } else {
                int level = 0;
                level = findout(arr);
                if (level == -1) {
                    ans = -1;
                } else {
                    ans = n + n - level - 1;
                }
            }
            out.println(ans);
        }

        int findout(int[] a) {
            int l = a.length;
            if (l == 1) {

                if (a[0] == 1) {
                    return l;
                } else {
                    return -1;
                }
            }
            boolean found = false;
            for (int i = 0; i < l; i++) {
                if (a[i] == 1) {
                    found = true;
                    break;
                }
            }
            if (found) {
                return l;
            } else {
                int[] arr1 = new int[l - 1];
                for (int i = 0; i <= l - 2; i++) {
                    arr1[i] = gcd(a[i], a[i + 1]);
                }
                return findout(arr1);
            }
        }

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

    }
}

