Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>. 

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2
1 1


Output

YES
1


Input

3
6 2 4


Output

YES
0


Input

2
1 3


Output

YES
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with <image>.

In the second example the gcd of the sequence is already greater than 1. 
import java.util.Scanner;

/**
 * Created by Kafukaaa on 2017/5/2.
 */
public class MikeAndGcdProblem {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        Integer n = scanner.nextInt();
        Integer[] s = new Integer[n];
        for (int i = 0; i < n; i++) {
            s[i] = scanner.nextInt();
        }

        Integer result = 0;
        Integer flag = gcd(s[0], s[1]);
        for (int i = 2; i < n; i++) {
            flag = gcd(s[i], flag);
        }
        if (flag > 1) {
            System.out.println("YES");
            System.out.println(0);
            System.exit(0);
        }

        for (int i = 0; i < n - 1; i++) {
            if (s[i] % 2 == 1 && s[i + 1] % 2 == 1) {
                s[i] = 2;
                s[i + 1] = 2;
                result++;
            }
        }

        for (int i = 0; i < n - 1; i++) {
            if ((s[i] % 2 == 0 && s[i + 1] % 2 == 1) || s[i] % 2 == 1 && s[i + 1] % 2 == 0) {
                s[i] = 2;
                s[i + 1] = 2;
                result += 2;
            }
        }
        System.out.println("YES");
        System.out.println(result);
    }

    private static Integer gcd(Integer a, Integer b) {
        if (a % b == 0) {
            return Math.abs(b);
        } else {
            return gcd(b, a % b);
        }
    }
}
