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.util.*;


public class Main {

    private static class CantDoException extends RuntimeException {

    }



    static class Task {
        PrintWriter out;


        int  gcd ( int a, int b )
        {
            int c;
            while ( a != 0 ) {
                c = a; a = b%a;  b = c;
            }
            return b;
        }

        public void solve(MyScanner in, PrintWriter out) {
            this.out = out;


            int n = in.nextInt();
            int[] a = new int[n + 22];

            int num1 = 0;
            int gcd = 0;
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
                if ( i == 0) {
                    gcd = a[i];
                } else {
                    gcd = gcd(gcd, a[i]);
                }

                if (a[i] == 1) {
                    num1++;
                }
            }

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

            if (num1 > 0) {
                out.println(n - num1);
                return;
            }

            int bestFind = Integer.MAX_VALUE;

            for (int i = 0; i < n - 1; i++) {
                int currentFind = 0;
                gcd = a[i];
                int index = i + 1;
                while (index < n && gcd != 1) {
                    gcd = gcd(gcd, a[index]);
                    index++;
                    currentFind++;
                }
                if (gcd == 1) {
                    bestFind = Math.min(bestFind, currentFind);
                }
            }

            out.println(n - 1 + bestFind);



        }


    }

    public static void main(String[] args) {
        MyScanner in = new MyScanner();
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}


