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.*;
public class C798 {
	public static long gcd(long a, long b){
		if(b==0)
			return a;
		return gcd(b,a%b);
	}
	public static long[] op(int i, long[] a){
		long temp1=a[i];
		long temp2=a[i+1];
		a[i]=temp1-temp2;
		a[i+1]=temp1+temp2;
		return a;
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		long[] a = new long[n];
		for(int i=0; i<n; i++) a[i]=s.nextLong();
		
		long g = gcd(a[0],a[1]);
		for(int i=2;i<n;i++){
			g=gcd(g,a[i]);
		}
		if(g>1){
			System.out.println("YES");
			System.out.println(0);
		}else{
			long count=0;
			
			for(int i=0;i<n-1;i++){
				while((a[i] & 1) == 1){
					a=op(i,a);
					count++;
				}
			}
			while(a[n-1]%2!=0){
				a=op(n-2,a);
				count++;
			}
			// 9 57 30 28 81 88 32 3 42 25

			
			System.out.println("YES");
			System.out.println(count);
			
		}
		
		
		
	}

}
