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.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            int n = in.nextInt();
            int[] arr = new int[n];
            arr[0] = in.nextInt();
            int[][] gcd = new int[n][n];
            gcd[0][0] = arr[0];
            int countOfOneInArr = arr[0] == 1 ? 1 : 0;
            int limit = n - 1;
            for (int i = 1; i < n; ++i) {
                arr[i] = in.nextInt();
                gcd[0][i] = getGCD(gcd[0][i - 1], arr[i]);
                if (i < limit && gcd[0][i] == 1) {
                    limit = i;
                }
                if (arr[i] == 1) countOfOneInArr++;
            }
            if (gcd[0][n - 1] > 1) {
                System.out.println("-1");
            } else if (countOfOneInArr > 0) {
                System.out.println(n - countOfOneInArr);
            } else {
                System.out.println(n - 1 + minDisWithGCDOne(arr, n, gcd, limit));
            }
        }
    }

    private static int minDisWithGCDOne(int[] arr, int n, int[][] gcd, int limit) {
        for (int i = 1; i < n; ++i) {
            gcd[i][i] = arr[i];
        }
        int before;
        for (int i = 1; i < limit; ++i) {
            for (int j = i + 1; j < n; ++j) {
                before = j - i;
                gcd[before][j] = getGCD(gcd[before][j - 1], arr[j]);
                if (gcd[before][j] == 1) {
                    return i;
                }
            }
        }
        return limit;
    }

    private static int getGCD(int a, int b) {
        int mod;
        while (b > 1) {
            mod = a % b;
            a = b;
            b = mod;
        }
        return b == 0 ? a : 1;
    }
}