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.*;
import java.io.*;

public class Pride {

	// https://codeforces.com/contest/891/problem/A
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("Pride"));

		int n = Integer.parseInt(in.readLine());
		int[] arr = new int[n];
		int ones=0;
		StringTokenizer st = new StringTokenizer(in.readLine());
		for (int i=0; i<n; i++) {
			arr[i] = Integer.parseInt(st.nextToken());
			if (arr[i] == 1) ones++;
		}
		
		if (ones > 0) {
			System.out.println(n - ones);
			return;
		}
		
		int gcd = arr[0];
		for (int i=1; i<n; ++i ) {
			gcd = gcd(gcd, arr[i]);
		}
		
		if (gcd != 1) {
			System.out.println(-1);
			return;
		}
		
		for (int i=1; i<n; i++ ) {
			if (gcd(arr[i], arr[i-1]) == 1) {
				System.out.println(n);
				return;
			}
		}

		int allmin=(int)(1e9);
		for (int i=0; i<n; i++) {
			int min=(int)(1e9);
			gcd = arr[i];
			for (int j=i-1; j>=0; j--) {
				gcd = gcd(gcd, arr[j]);
				if (gcd == 1) {
					min = i - j + (n-1);
					break;
				}
			}
			gcd = arr[i];
			for (int j=i+1; j<n; j++) {
				gcd = gcd(gcd, arr[j]);
				if (gcd == 1) {
					min = Math.min(min, j - i + (n-1));
					break;
				}
			}
			allmin = Math.min(allmin, min);
		}
		
		System.out.println(allmin);
	
	}
	
	public static int gcd(int a, int b) { 
        if (b == 0) return a; 
        return gcd(b, a%b); 
    } 
}