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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class CF13 {

	final static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	final static PrintWriter pw = new PrintWriter(System.out);

	public static void main(String[] args) {

		int n = readInt();
		int[] a = readIntArr();
		if (n == 1) {

			if (a[0] == 1) {
				pw.println(0);
			} else {
				pw.println(-1);
			}
		} else {
			int nm1 = 0;

			for (int i = 0; i < n; i++) {
				if (a[i] == 1) {
					nm1++;
				}
			}
			if (nm1 == 0) {

				int min = Integer.MAX_VALUE;
				for (int i = 0; i < n; i++) {

					if (i < n - 1) {
						int gcdV = a[i];
						int nm = 0;
						for (int j = i + 1; j < n; j++) {
							gcdV = gcd(gcdV, a[j]);
							nm++;
							if (gcdV == 1) {
								min = Math.min(nm + n - 1, min);

								break;
							}
						}
						gcdV = a[i];
						nm = 0;
						for (int j = i - 1; j >= 0; j--) {
							gcdV = gcd(gcdV, a[j]);
							nm++;
							if (gcdV == 1) {
								min = Math.min(nm + n - 1, min);

								break;
							}
						}

					}
				}
				if (min == Integer.MAX_VALUE) {
					pw.println(-1);
				} else {
					pw.println(min);
				}
			} else {
				pw.println(n - nm1);
			}
		}
		pw.close();

	}

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

	static String readLine() {
		String line = null;
		try {
			line = br.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return line;

	}

	static String readString() {
		return readLine();

	}

	static public long readlong() {
		return Long.parseLong(readLine());
	}

	static public int readInt() {
		return Integer.parseInt(readLine());
	}

	static String[] stringArray() {

		StringTokenizer st = new StringTokenizer(readLine());
		int n = st.countTokens();
		String ret[] = new String[n];
		for (int i = 0; i < n; i++) {
			ret[i] = st.nextToken();
		}
		return ret;

	}

	static public int[] readIntArr() {
		String[] str = stringArray();
		int arr[] = new int[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Integer.parseInt(str[i]);
		return arr;
	}

	static public double[] readDoubleArr() {
		String[] str = stringArray();
		double arr[] = new double[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Double.parseDouble(str[i]);
		return arr;
	}

	static public long[] readLongArr() {
		String[] str = stringArray();
		long arr[] = new long[str.length];
		for (int i = 0; i < arr.length; i++)
			arr[i] = Long.parseLong(str[i]);
		return arr;
	}

	static public double readDouble() {
		return Double.parseDouble(readLine());
	}
}