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.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		
		int[] arr=new int[n];
		
		for(int i=0;i<n;i++)
		{
			arr[i]=s.nextInt();
		}
		
		int gcd=arr[0];
		
		for(int i=1;i<n;i++)
		{
			gcd=gcd(gcd,arr[i]);
		}
		
		if(gcd>1)
		{
			System.out.println("YES");
			System.out.println(0);
		}
		else
		{
			int count=0;
			
			for(int i=0;i<n-1;i++)
			{
				if(arr[i]%2==1&&arr[i+1]%2==1)
				{
					arr[i]=2;
					arr[i+1]=2;
					
					count++;
				}
			}
			
			for(int i=0;i<n-1;i++)
			{
				if(arr[i]%2!=arr[i+1]%2)
				{
					arr[i]=2;
					arr[i+1]=2;
					count=count+2;
				}
			}
			
			System.out.println("YES");
			
			System.out.println(count);
		}
		
		
		
	}
	
	public static int gcd(int a,int b)
	{
		if(b==0)
			return a;
		
		return gcd(b,a%b);
	}
	
}