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.*;
import static java.lang.System.out;
/**
 * @date2017年4月22日下午11:14:40
 */
public class C798 {
    /**
     * @param args
     */
    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    public static void main(String[] args) {
        @SuppressWarnings("resource")
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int con = 0;
        int ans = 0;
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            if ((a[i] & 1) == 1) {
                con++;
            } else {
                if (con != 0) {
                    ans += (con >> 1) + ((con & 1) << 1);
                    con = 0;
                }
            }
        }
        if (con != 0)
            ans += (con >> 1) + ((con & 1) << 1);            
        out.println("YES");
        
        con = a[0];
        for (int i = 1; i < n; i++) {
            con = gcd(con, a[i]);
            if (con == 1){
                out.println(ans);
                return;
            }
        }
        out.println(0);
    }
}