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.Arrays;
import java.util.StringTokenizer;

/**
 *
 * @author arif_
 * date:
 * algo:
 * diff:
 */
public class CF_892C {
    /* START OF I/O ROUTINE */
    // PrintWriter for faster output
    public static PrintWriter out;

    // MyInputReader class for faster input
    public static class MyInputReader {
        BufferedReader br;
        StringTokenizer st;

        public MyInputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
        }

        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;
        }
    } // end of class MyInputReader
    /* END OF I/O ROUTINE */

    static int[][] dp; // dp[i][j] contains gcd(a[i]...a[j]
    static int[] a;


    private static int rangeGCD(int i, int j) {
        if (dp[i][j] != 0)
            return dp[i][j];

        if (i+1 == j)
            return dp[i][j] = gcd(a[i], a[j]);

        return dp[i][j] = gcd(rangeGCD(i, j-1), a[j]);
    }


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


    public static void main(String[] args) {
        MyInputReader in = new MyInputReader(System.in);
        out = new PrintWriter(new BufferedOutputStream(System.out));

        int n = in.nextInt();

        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }

        int ans = -1;

        int ones = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] == 1) ones++;
        }
        if (ones > 0) {
            ans = n - ones;
        }

        if (ans == -1) {
            dp = new int[n+2][n+2];
            for (int range=2; range<=n; range++) {
                for (int lt = 0, rt = lt + range - 1; rt < n; lt++, rt++) {
                    if (1 == rangeGCD(lt, rt)) {
                        //out.println(range + " " + lt + " " + rt);
                        ans = (range - 1) + (n - 1);
                        break;
                    }
                }
                if (ans != -1) break;
            }
        }

        out.println(ans);
        out.close();
    } // end of method main()
} // end of class Main