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.*;
import java.text.*;
import java.math.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Map.Entry;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskC {
		static int gcd(int x, int y) {
			return x == 0 ? y : gcd(y % x, x);
		}

		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int[] a = new int[n];
			int oneC = 0;
			for (int i = 0; i < n; i++) {
				a[i] = in.nextInt();
				if (a[i] == 1) {
					oneC++;
				}
			}
			if (oneC > 0) {
				out.println(n - oneC);
				return;
			}
			int minC = n + 2;
			for (int i = 0; i < n; i++) {
				int c = 0;
				int v = a[i];
				int j = i + 1;
				while (j < n && v != 1 && minC > c) {
					v = gcd(v, a[j]);
					c++;
					j++;
				}
				if (v == 1) {
					minC = c;
				}
			}
			for (int i = n - 1; i >= 0; i--) {
				int c = 0;
				int v = a[i];
				int j = i - 1;
				while (j >= 0 && v != 1 && minC > c) {
					v = gcd(v, a[j]);
					c++;
					j--;
				}
				if (v == 1) {
					minC = c;
				}
			}
			if (minC == n + 2) {
				out.println(-1);
				return;
			}
			out.println(minC + n - 1);
		}
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

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

		public double nextFloat() {
			return Double.parseDouble(next());
		}

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

	}
}
