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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class A {

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

	public static void main(String[] args) throws NumberFormatException, IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt();

		int[] a = new int[n];
		
		int cc = 0;

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			if(a[i] == 1)
				cc++;
		}
		
		if(cc > 0) {
			out.println(n - cc);
			out.flush();
			return;
		}

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

		out.flush();
		out.close();
	}


	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}
		public Scanner(String file) throws FileNotFoundException{	br = new BufferedReader(new FileReader(file));}


		public String next() throws IOException {
			while(st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}
		public long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

	}

}

