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.io.*;
import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		PrintWriter pw=new PrintWriter(System.out);
		while(sc.hasNext()){
			int n=sc.nextInt();
			int[]shu=new int[n];
			for(int i=0;i<n;i++) shu[i]=sc.nextInt();
			long g=gcd(shu[0],shu[1]);
			for(int i=2;i<n;i++){
				g=gcd(g,shu[i]);
			}
			if(g>1){
				pw.println("YES");
				pw.println(0);
			}else{
				ArrayList<Integer>list=new ArrayList<Integer>();
				boolean flag=true;
				int min=0;
				  for(int i=0;i<n;i++){
					  if(shu[i]%2!=0){
						  list.add(shu[i]);
					  }else{
						  if(list.size()%2==0){
							  min+=list.size()/2;
						  }else{
							  min+=list.size()/2+2;
						  }
						  list.clear();
					  }
				  }
				  if(list.size()>0){
					  if(list.size()%2==0){
						  min+=list.size()/2;
					  }else{
						  min+=list.size()/2+2;
					  }
				  }
					  pw.println("YES");
					  pw.println(min);
			}
			pw.flush();
		}
	}
	public static long gcd(long a,long b){
		return a == 0 ? b : gcd(b % a, a);
	}
}