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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;


public class A {
    public static int gcd(int a, int b){
        if (b == 0) return a;
        return gcd(b, a % b);
    }
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
//        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
//        String[] ar = scan.readLine().split(" ");
        int n = in.nextInt();
        int a[] = new int[n + 1];
        int d[][] = new int[n + 1][n + 1];
        int b[][] = new int[n + 1][n + 1];
        for (int i = 1;i <= n; i++) {
            a[i] = in.nextInt();
            b[i][i] = a[i];
        }
        for (int k = 1; k <= n; k++)
            for (int j = 1; j <= n - k; j++) {
                int i = j + k;
                int p = gcd(b[j][i - 1], a[i]);
                int q = gcd(b[j + 1][i], a[j]);
                int t1 = (b[j][i - 1] > 1 ? 1 : 0) + (a[i] > 1 ? 1 : 0);
                int t2 = (b[j + 1][i] > 1 ? 1 : 0) + (a[j] > 1 ? 1 : 0);
//                System.out.println(j + " " + i + " " + p + " " + q);
                if (p == 1 && q == 1) {
                    d[j][i] = Math.min(d[j + 1][i] + t2, d[j][i - 1] + t1);
                    b[j][i] = 1;
                } else
                if (p == 1) {
                    d[j][i] = d[j][i - 1] + t1;
                    b[j][i] = 1;                
                } else 
                if (q == 1)
                {
                    d[j][i] = d[j + 1][i] + t2;
                    b[j][i] = 1;                                
                } else {
                    if (d[j + 1][i] > d[j][i - 1]) {
                        d[j][i] = d[j][i - 1] + t1;
                        b[j][i] = p;
                    } else {
                        d[j][i] = d[j + 1][i] + t2;
                        b[j][i] = q;
                    }
                }
            }
//        for (int i = 1; i <= n; i++) {
//            for (int j = 1; j <= n; j++)
//                System.out.print(d[i][j] + " ");
//            System.out.println();
//        }
//        System.out.println("-----");
//        for (int i = 1; i <= n; i++) {
//            for (int j = 1; j <= n; j++)
//                System.out.print(b[i][j] + " ");
//            System.out.println();
//        }
        if (b[1][n] == 1) {
            System.out.println(d[1][n]);
        } else {
            System.out.println(-1);
        }
    }
}