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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class CF891A {
	static int count = 0;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		List<Integer> list = new ArrayList<>();
		int i = 0;
		while (i++ < n) {
			list.add(sc.nextInt());
		}
		int gcd = list.get(0);
		if (list.contains(1)) {
			gcd = 1;
			for (i = 0; i < list.size(); i++) {
				if (list.get(i) != 1) {
					count++;
				}
			}
		} else {

			for (i = 1; i < list.size(); i++) {

				if (gcd != list.get(i)) {
					gcd = gcd(gcd, list.get(i));
					count++;
				}
				if (gcd == 1) {
					count += list.size() - 1;
					break;
				}

			}
		}
		int ans2 = 999999999;
		int gcd2 = 0;
		for (i = 0; i < list.size() - 1; i++) {
			gcd2 = gcd(list.get(i), list.get(i + 1));
			if (gcd2 == 1) {
				ans2 = n;
				break;
			}
		}

		if (gcd != 1) {
			count = -1;
		}
		count = Math.min(count, ans2);

		System.out.println(count);

	}

	private static int gcd(int a, int b) {
		while (b > 0) {
			int temp = b;
			b = a % b; // % is remainder
			a = temp;
		}
		return a;
	}

}
