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.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.valueOf(st.nextToken());
		int a[] = new int[n + 1];

		st = new StringTokenizer(br.readLine());
		int cnt = 0;
		for (int i = 1; i <= n; i++) {
			a[i] = Integer.valueOf(st.nextToken());
			if (a[i] == 1)
				cnt++;
		}

		if (cnt > 0) {
			System.out.println(n - cnt);
			return;
		}

		for (int i = 1; i <= n - 1; i++) {
			int mmax = Math.max(a[i], a[i + 1]);
			int mmin = Math.min(a[i], a[i + 1]);
			if (mmax == mmin)
				continue;
			if (1 == gcd(mmax, mmin)) {
				System.out.println(n);
				return;
			}
		}

		int minLen = Integer.MAX_VALUE;

		for (int i = n - 1; i >= 1; i--) {
			int tmp = a[i];
			int cntOp = 0;
			for (int j = i; j <= n; j++) {
				if (a[j] == a[i])
					continue;
				int mmax = Math.max(tmp, a[j]);
				int mmin = Math.min(tmp, a[j]);
				int gmod = gcd(mmax, mmin);
				//System.out.println("gmod==" + gmod);
				if (gmod == a[j])
					continue;
				cntOp++;
				tmp = gmod;
				if (gmod == 1) {
					minLen = Math.min(minLen, cntOp);
					break;
				}
			}
		}

		if (minLen == Integer.MAX_VALUE)
			System.out.println(-1);
		else {
			System.out.println(minLen + n - 1);
		}

	}

	static public int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
}