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.*;
import java.util.*;

public class Main {
	static int[] v;
	static int[] tree;
	static final int INF = (int)1e9 + 5;

	static void build(int node, int a, int b) {
		if(a == b) {
			tree[node] = v[a];
			return;
		}
		int mid = (a + b) / 2;
		build(node * 2, a, mid);
		build(node * 2 + 1, mid + 1, b);
		tree[node] = gcd(tree[node * 2], tree[node * 2 + 1]);
		return;
	}

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

	static int query(int node, int a, int b, int i, int j) {
		if (a > b || b < i || a > j) {
			return 0;
		}
		if (i <= a && b <= j) {
			return tree[node];
		}
		int mid = (a + b) / 2;
		int q1 = query(node * 2, a, mid, i, j);
		int q2 = query(node * 2 + 1, mid + 1, b, i, j);
		return gcd(q1, q2);
	}
		
    public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		v = new int[n + 1];
		tree = new int[4 * n + 5];
	
		int qt_one = 0;
		for (int i = 1; i <= n; i++) {
			v[i] = sc.nextInt();
			if (v[i] == 1) {
				qt_one++;
			}
		}

		int res = INF;
		if (qt_one > 0) {
			System.out.println(n - qt_one);
		} else {
			build(1, 1, n);
			for (int i = 1; i <= n; i++) {
				for (int j = i + 1; j <= n; j++) {
					int d = query(1, 1, n, i, j);
					if (d == 1) {
						res = Math.min(res, j - i + n - 1);
					}
				}
			}

			if (res < INF) {
				System.out.println(res);
			} else {
				System.out.println(-1);
			}
		}	
    }
};
// 1511547269022
