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.*;
import java.math.*;
public class Main {
	static Scanner cin = new Scanner(System.in);
	static PrintWriter cout = new PrintWriter(System.out);
	static int gcd(int a, int b) {
		if (b == 0) return a;
		return gcd(b, a % b);
	}
	static int[] a = new int[2010];
	public static void main(String[] args) {
		int n = cin.nextInt();
		for (int i = 1; i <= n; ++ i) a[i] = cin.nextInt();
		int d = a[1];
		for (int i = 2; i <= n; ++ i) d = gcd(d, a[i]);
		if (d != 1) {
			cout.println(-1);
			cout.flush();
			System.exit(0);
		}
		int One = 0;
		for (int i = 1; i <= n; ++ i) 
		if (a[i] == 1) One ++;
		if (One > 0) {
			cout.println(n - One);
			cout.flush();
			System.exit(0);
		}
		int dis = n;
		for (int i = 1; i <= n; ++ i) {
			int td = a[i];
			if (td == 1) {
				dis = Math.min(dis, td);
			}
			else {
				for (int j = i + 1; j <= n; ++ j) {
					td = gcd(td, a[j]);
					if (td == 1) {
						dis = Math.min(dis, j - i);
						break;
					}
				}
			}
		}
		cout.println(n - 1 + dis);
		cout.flush();
	}
}