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 tr {
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int []a=new int [n];
		int no=0;
		for(int i=0;i<n;i++) {
			a[i]=sc.nextInt();
		    if(a[i]!=1)
		    	no++;
		}
		if(n==1) {
			if(a[0]==1)
				System.out.println(0);
			else
				System.out.println(-1);
			return;
		}
		for(int i=0;i<n-1;i++) {
			if(gcd(a[i],a[i+1])==1) {
				System.out.println(no);
				return;
			}
		}
		int ans=Integer.MAX_VALUE;
		boolean ff=false;
		for(int i=0;i<n;i++) {
			int ti=0;
			int d=a[i];
			for(int j=i+1;j<n;j++) {
				d=gcd(d,a[j]);
				if(d==1) {
					ans=Math.min(ans, j-i);
					ff=true;
					break;
				}
			}
		}
		if(ff)
			out.print(ans+n-1);
		else
			out.println(-1);
		out.flush();
	}

	static int tryy(int[] a) {
		if (a.length == 1)
			return 1;
		if (issorted(a))
			return a.length;
		int[] a1 = new int[a.length / 2];
		int[] a2 = new int[a.length / 2];
		int j = 0;
		for (int i = 0; i < a.length; i++) {
			if (i < a.length / 2)
				a1[i] = a[i];
			else {
				a2[j++] = a[i];
			}
		}
		if (issorted(a1) || issorted(a2))
			return a1.length;
		return Math.max(tryy(a1), tryy(a2));
	}

	static boolean issorted(int[] a) {
		for (int i = 0; i < a.length - 1; i++)
			if (a[i] > a[i + 1])
				return false;
		// System.out.println(Arrays.toString(a));
		return true;
	}

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

	static class op {
		int a;
		int b;

		op(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public String toString() {
			return a + " " + b;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			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 IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

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

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}