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 Solution
{
    static  int gcd(int a, int b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
} 
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
            int n = sc.nextInt();
             int arr[] = new int[n];
             arr[0] = sc.nextInt();
             int g = arr[0];
             for(int i = 1 ; i < n ; i++)
             {
                    arr[i] = sc.nextInt();
                    g = gcd(arr[i] , g);
             }
             
             int i = 0;
             int ans = 0;
            
             while(i < n)
             {
                    int j = i;
                    while(j+1 < n && arr[j+1]%2 == arr[j]%2)
                    {
                           j++;
                    }
                    if(arr[i]%2 == 1)
                    {
                        if((j-i+1)%2 == 1)
                        {
                        ans += (j-i+1)/2 + 2;
                        }
                        
                        else
                        ans += (j-i+1)/2;
                    }
                    i = j+1;
             }
            
             System.out.println("YES");
             if(g != 1)
             System.out.println(0);
             
             else
             System.out.println(ans);
             
       }
}