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.
// package codeforces.A;

import java.util.*;

public class cf891A {

    private static int gcd(int x, int y) {
        if (0 == y) return x;
        return gcd(y, x % y);
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int n = in.nextInt();
        int[] numbers = new int[n];
        int g = 0;
        // 计算输入中是否含有一
        int count = 0;
        for (int i = 0; i < n; i++) {
            numbers[i] = in.nextInt();
            if (numbers[i] == 1) {
                count++;
            }
            g = gcd(g, numbers[i]); // 计算所有的输入中是否含有同样的公约数
        }

        if (count > 0) {
            System.out.print(n-count);
            return;
        }
        // 如果g大于1，说明不能够将所有的数字都变成1
        if (g > 1) {
            System.out.print(-1);
            return;
        }

        int move = (int) 1e7;
        for (int i = 0; i < n; i++) {
            int common = 0;
            for (int j = i; j < n; j++) {
                common = gcd(common, numbers[j]);
                if (common == 1) {
                    // 将一个元素变成1的最小步数
                    move = Math.min(move, j - i);
                    break;
                }
            }
        }

        move += n - 1;

        System.out.print(move);
    }
}
