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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class C_Pride {
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));

    public static void main(String[] args) throws Exception {
        int n = Integer.valueOf(rd.readLine());
        String[] ar = rd.readLine().split(" ");
        int[] v= new int[n];

        int gcd = 0;
        for (int i = 0; i < n; i++) {
            v[i] = Integer.valueOf(ar[i]);
            gcd = gcd(gcd, v[i]);
        }
        if (gcd != 1){
            System.out.println(-1);
            return;
        }
        int ans = 0;
        for (int i = 0; i < n; i++){
            if (v[i] == 1){
                for (int j = 0; j < n; j++) {
                    if (v[j] > 1) ans ++;
                }
                System.out.println(ans);
                return;
            }
        }

        ans = 100000000;
        for (int i = 0; i < n; i++) {
            gcd = 0;
            for (int j = i; j < n; j++) {
                gcd = gcd(gcd, v[j]);
                if (gcd  == 1){
                    ans = Math.min(ans, n - 1 + j - i);
                }
            }
        }
        System.out.println(ans);
    }

    static int gcd(int a, int b){
        if (a == 0 || b == 0) return a + b;
        if (a < b) return gcd(b, a);
        if (a % b == 0) return b;
        else return gcd(b, a % b);
    }
}
