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 java.math.*;
import java.io.*;
import java.text.*;

public class A{
static class Node implements Comparable<Node> {
	int id;
	long cost;
	public Node(int id,long cost) {
		this.id=id;
		this.cost=cost;
	}
	public int compareTo(Node c) {
		return Long.compare(this.cost, c.cost);
	}
}
//public static PrintWriter pw; 	
public static PrintWriter pw=new PrintWriter(System.out);
public static void solve() throws IOException{
//	pw=new PrintWriter(new FileWriter("C:\\Users\\shree\\Downloads\\small_output_in"));
	FastReader sc=new FastReader();
	int n=sc.I();
	long a[]=new long[n+1];
	long d=0;
	for(int i=1;i<=n;i++) {
		a[i]=sc.L();
		d=gcd(a[i],d);
	}
	 if(d>1) {
		 pw.println("YES\n0");
		 pw.close();
		 return;
	 }
	 int cnt=0;
	 for(int i=1;i<n;i++) {
		 if(a[i]%2!=0) {
			 long a1=a[i]; long a2=a[i+1];
			 a[i]=a1-a2;
			 a[i+1]=a1+a2;
			 if(a2%2==0) {
				a1=a[i]; a2=a[i+1];
				 a[i]=a1-a2;
				 a[i+1]=a1+a2;
				 cnt+=2;
			 }else cnt++;
		 }
	 }
	 if(a[n]%2!=0) cnt+=2;
	
		 pw.println("YES\n"+cnt);
	 
	
	 
	 pw.close(); 
	
	
}
static long gcd(long a,long b) {
	if(b==0) return a;
	return gcd(b,a%b);
}
public static void main(String[] args)  {
    new Thread(null ,new Runnable(){
      public void run(){
          try{
          
              solve();
          } catch(Exception e){
              e.printStackTrace();
          }
      }
  },"1",1<<26).start();
 
}

static long M=(long)Math.pow(10,9)+7;
 
	static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader() throws FileNotFoundException{
            //br=new BufferedReader(new FileReader("C:\\Users\\shree\\Downloads\\B-small-practice.in"));	
        	 br = new BufferedReader(new InputStreamReader(System.in));
        		
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
       
        int I(){   return Integer.parseInt(next()); }
        long L(){  return Long.parseLong(next()); }
        double D() {  return Double.parseDouble(next()); }
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
     }
}