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
 *
 * @author Aeroui
 */
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 gcd = 0;

            int[] arr = new int[n + 1];

            int[][] dp = new int[n + 1][n + 1];
            boolean flag = false;
            for (int i = 1; i <= n; ++i) {
                arr[i] = in.nextInt();
                gcd = gcd(arr[i], gcd);

                dp[i][i] = arr[i];
                if (arr[i] == 1) {
                    flag = true;
                }
            }

            if (gcd != 1) {
                out.println(-1);
                return;
            }

            if (flag) {
                int ans = 0;
                for (int i = 1; i <= n; ++i) {
                    if (arr[i] != 1)
                        ++ans;
                }

                out.println(ans);

            } else {
                int min = Integer.MAX_VALUE;
                for (int i = 1; i <= n; ++i) {
                    for (int j = i + 1; j <= n; ++j) {
                        dp[i][j] = gcd(dp[i][j - 1], arr[j]);
                        if (dp[i][j] == 1) {
                            min = Math.min(min, j - i + 1);
                        }
                    }
                }

                out.println(n - 1 + (min - 1));
            }

        }

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

    }
}