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 
{
    public static void main(String[] args) 
    {
        // TODO code application logic here
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int[] arr = new int[n];
        for(int i = 0 ; i < n ; i++)
            arr[i] = input.nextInt();
        boolean check = true;
        for(int i = 0 ; i < n ; i++)
        {
            if(arr[i] != 0)
            {
                check = false;
                break;
            }
        }
        if(check)
            System.out.println("NO");
        else
        {
            int gcd = arr[0];
            for(int i = 1 ; i < n ; i++)
            {
                gcd = gcd(gcd,arr[i]);
            }
            if(gcd > 1)
                System.out.println("YES\n0");
            else
            {
                //make all the numbers divisble by 2
                int count = 0;
                for(int i = 1 ; i < n ; i++)
                {
                    if(arr[i - 1] % 2 == 1 && arr[i] % 2 == 1)
                    {
                        count++;
                        arr[i] = 0;                        
                    }
                    else if(arr[i - 1] % 2 == 1)
                    {
                        count += 2;
                        arr[i] = 0;
                    }
                }
                if(arr[n - 1] % 2 == 1)
                    count +=2;
                System.out.println("YES\n" + count);
            }
        }
    }
    public static int gcd(int a , int b)
    {
        if(b == 0)
            return a;
        return gcd(b,a%b);
    }
}