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.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;

public class Main {
	
	static int[] arr;
	
	public static void main(String[] args) {
		FastScanner input = new FastScanner(System.in);
		
		int N = input.nextInt();
		
		int[] arr = new int[N];
		for (int i = 0; i < N; i++)
			arr[i] = input.nextInt();
		
		int cnt = 0;
		for (int i = 0; i < N; i++)
			if (arr[i] == 1)
				cnt++;
		
		if (cnt > 0)
			System.out.println(N - cnt);
		else {
			int min = Integer.MAX_VALUE;
			for (int i = 0; i < N; i++) {
				int gcd = arr[i];
				for (int j = i; j < N; j++) {
					gcd = gcd(gcd, arr[j]);
					if (gcd == 1) {
						min = Math.min(min, j - i + 1);
						break;
					}
				}
			}
			if (min == Integer.MAX_VALUE)
				System.out.println(-1);
			else
				System.out.println(min + N - 2);
		}
	}
	
	static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}
	
	// Matt Fontaine's Fast IO
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

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

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

		String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}
}
