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 scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] arr = new int[n];
        int check = 0;
        int count = 0;
        for(int i = 0; i < n; i++)
        {
            arr[i] = scan.nextInt();
            if(i != 0)
                check = gcd(check, arr[i]);
            else
                check = arr[i];
        }

        if(check > 1)
            System.out.println("YES\n0");
        else
        {
            for(int i = 0; i < n - 1; i++)
            {
                if(arr[i] % 2 == 1 && arr[i+1] % 2 == 1)
                {
                    count++;
                    arr[i] = 2;
                    arr[i+1] = 2;
                }
            }
            for(int i = 0; i < n - 1; i++)
            {
                if((arr[i] % 2 == 1 && arr[i+1] % 2 == 0) || (arr[i] % 2 == 0 && arr[i+1] % 2 == 1))
                {
                    count += 2;
                    arr[i] = 2;
                    arr[i+1] = 2;
                }
            }

            System.out.println("YES\n" + count);
        }
    }

    static int gcd(int x, int y)
    {
        return y == 0 ? x : gcd(y, x % y);
    }
}

	 	 		  	 		 	  	 	 					 		 		