Problem: Acacius is studying strings theory. Today he came with the following problem.

You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?

Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".

Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 ≤ i ≤ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.

Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.

Input

First line of input contains an integer T (1 ≤ T ≤ 5000), number of test cases. T pairs of lines with test case descriptions follow.

The first line of a test case description contains a single integer n (7 ≤ n ≤ 50), length of a string s.

The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.

Output

For each test case output an answer for it.

In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".

Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.

You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).

Example

Input


6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba


Output


Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No

Note

In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.

In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.

In sixth example there are two occurrences of a string "abacaba" as a substring.

Solution (in JAVA):

Solution:
/**
 * @author vivek
 * programming is thinking, not typing
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

//todo
public class Main {

    private static void solveTC(int __) {

        /* For Google */
//		ans.append("Case #").append(__).append(": ");

        //code start todo

        int n = Integer.parseInt(scn.nextLine());

        String target="abacaba";
        String str=scn.nextLine();

        int count=0;
        for (int i = 0; i <= n-7; i++) {
            if (str.substring(i).startsWith(target)){
                count++;
            }
        }

        if (count>1){
            print("No\n");
            return;
        }
        if (count==1){
            print("Yes\n");
            print(toOk(str,n)+"\n");
            return;
        }

        ArrayList<String> list=new ArrayList<>();

        outer:
        for (int i = 0; i <=n-7 ; i++) {
            String temp=str.substring(i,i+7);
            for (int j = 0; j < 7; j++) {
                if (temp.charAt(j)==target.charAt(j)||temp.charAt(j)=='?'){
                }else {
                    continue outer;
                }
            }
            list.add(toOk(str.substring(0,i),i)+target+toOk(str.substring(i+7),n-i-7));
        }

        for(String ele:list){
            int count2=0;
            for (int i = 0; i <= n-7; i++) {
                if (ele.substring(i).startsWith(target)){
                    count2++;
                }
            }

            if (count2==1){
                print("Yes\n");
                print(ele+"\n");
                return;
            }
        }

        print("No\n");

        //code end

    }

    private static String toOk(String str,int n) {
        StringBuilder ans= new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (str.charAt(i)=='?'){
                ans.append('d');
            }else {
                ans.append(str.charAt(i));
            }
        }
        return ans.toString();
    }

    /**
     * Don't copy other's templates, make your own <br>
     * It would be much more beneficial
     */

    public static void main(String[] args) {

//        int limit=               ;
//        sieve(limit);

        //todo
/*
        try {
            System.setOut(new PrintStream(new File("file_i_o\\output.txt")));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
*/

        scn = new Scanner();
        ans = new StringBuilder();
        int t = Integer.parseInt(scn.nextLine());
//		int t = 1;
        for (int i = 1; i <= t; i++) {
            solveTC(i);
        }
        System.out.print(ans);
    }
    //Stuff for prime start

    /**
     * List containing prime numbers <br>
     * <b>i<sup>th</sup></b> position contains <b>i<sup>th</sup></b> prime number <br>
     * 0th index is <b>null</b>
     */
    private static ArrayList<Integer> listOfPrimes;

    /**
     * query <b>i<sup>th</sup></b> element to get if its prime of not
     */
    private static boolean[] isPrime;

    /**
     * Performs Sieve of Erathosnesis and initialise isPrime array and listOfPrimes list
     *
     * @param limit the number till which sieve is to be performed
     */
    private static void sieve(int limit) {
        listOfPrimes = new ArrayList<>();
        listOfPrimes.add(null);
        boolean[] array = new boolean[limit + 1];

        Arrays.fill(array, true);
        array[0] = false;
        array[1] = false;

        for (int i = 2; i <= limit; i++) {
            if (array[i]) {
                for (int j = i * i; j <= limit; j += i) {
                    array[j] = false;
                }
            }
        }
        isPrime = array;
        for (int i = 0; i <= limit; i++) {
            if (array[i]) {
                listOfPrimes.add(i);
            }
        }
    }
    //stuff for prime end

    /**
     * Calculates the Least Common Multiple of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Least Common Multiple of <b>a</b> and <b>b</b>
     */
    private static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    /**
     * Calculates the Greatest Common Divisor of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Greatest Common Divisor of <b>a</b> and <b>b</b>
     */
    private static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void print(Object obj) {
        ans.append(obj.toString());
    }


    static Scanner scn;
    static StringBuilder ans;

    //Fast Scanner
    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
            //todo
/*
            try {
                br = new BufferedReader(new
                        InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
*/
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] nextIntArray(int n) {
            int[] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        Integer[] nextIntegerArray(int n) {
            Integer[] array = new Integer[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        long[] nextLongArray(int n) {
            long[] array = new long[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextLong();
            }
            return array;
        }

        String[] nextStringArray() {
            return nextLine().split(" ");
        }

        String[] nextStringArray(int n) {
            String[] array = new String[n];
            for (int i = 0; i < n; i++) {
                array[i] = next();
            }
            return array;
        }
    }
}