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. 

/**
 * @author: Mehul Raheja
 */
import java.util.*;
import java.io.*;

public class mikegcd {

    /*
        Runtime = O()
     */
    static int N, M, K;
    static String s;
    static StringTokenizer st;


    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int N = Integer.parseInt(br.readLine());
        long[] d = new long[N];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            d[i] = Integer.parseInt(st.nextToken());
        }
        long gcd = d[0];
        for (int i = 0; i < N; i++) {
            gcd = gcd(gcd,d[i]);
        }
        
        if(gcd > 1){
            System.out.println("YES");
            System.out.println("0");
        }else{
            int c = 0;
            boolean[] even = new boolean[N];
            for (int i = 0; i < N; i++) {
                even[i] = (d[i]%2==0);
            }
            //System.out.println(Arrays.toString(even));
            for (int i = 0; i < even.length; i++) {
                if(even[i]){
                    continue;
                }
                
                boolean around = false;
                
                if(i < even.length-1){
                    if(!even[i+1]){
                        around = true;
                    }
                }
                
                if(around){
                    even[i] = true;
                    even[i+1] = true;
                    c++;
                }else{
                    c+=2;
                    even[i] = true;
                    
                }
            }
            System.out.println("YES");
            System.out.println(c);
        }
        
        
    }

    public static long gcd(long a, long b) {
        while (b > 0) {
            long t = b;
            b = a % b;
            a = t;
        }
        return a;
    }
}
