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 MikeAndGCDProblem {
    public static int gcd(int a,int b)
    {
        if(b==0)
        {
            return a;
        }
        else{
            return gcd(b,a%b);
        }
        
    }
    
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        
        int n=sc.nextInt();
        int result=0;
        int ans=0;
        int con=0;
        
        for(int i=0;i<n;i++) {
          int x=sc.nextInt();
          result=gcd(x,result);
          int check=x&1;
          if(check == 1){
            con++;
          }
          else {
            ans+=(con+1)/2+ (con&1);
            con=0;
          }
        }
        ans+=(con+1)/2+ (con&1);
        System.out.println("YES");
        if (result==1) {
          System.out.println(ans);    
        } else {
          System.out.println (0);
        }
     }      
}
    
