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:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        AAcaciusAndString solver = new AAcaciusAndString();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class AAcaciusAndString {
        private static final String CONSTANT = "abacaba";
        private static final int LEN = CONSTANT.length();

        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int n = in.nextInt();
            char[] s = in.nextString().toCharArray();
            String ss = new String(s);
            int[] occur = KMPMatcher.findAllOccurrences(new String(s), CONSTANT);
            if (occur.length == 0) {
                String ans = null;

                for (int i = 0; i + LEN <= n; ++i) {
                    String subs = ss.substring(i, i + LEN);
                    boolean poss = true;
                    for (int j = 0; j < LEN; ++j) {
                        if (subs.charAt(j) != '?' && subs.charAt(j) != CONSTANT.charAt(j)) {
                            poss = false;
                            break;
                        }
                    }
                    if (poss) {
                        String newString = ss.substring(0, i) + CONSTANT + ss.substring(i + LEN);
                        if (KMPMatcher.findAllOccurrences(newString, CONSTANT).length == 1) {
                            ans = fillZ(newString);
                            break;
                        }
                    }
                }
                if (ans == null) {
                    out.println("No");
                } else {
                    out.println("Yes");
                    out.println(ans);
                }
            } else if (occur.length == 1) {
                s = fillZ(s);
                out.println("Yes");
                out.println(s);
            } else {
                out.println("No");
            }
        }

        private char[] fillZ(char[] s) {
            for (int i = 0; i < s.length; ++i) {
                if (s[i] == '?') {
                    s[i] = 'd';
                }
            }
            return s;
        }

        private String fillZ(String ss) {
            return new String(fillZ(ss.toCharArray()));
        }

    }

    static class KMPMatcher {
        public static int[] kmpTable(char[] str) {
            int n = str.length;
            int[] kmp = new int[n + 1];
            kmp[0] = -1;
            kmp[1] = 0;
            for (int i = 2, j = 0; i <= n; i++) {
                while (j > 0 && str[i - 1] != str[j]) j = kmp[j];
                kmp[i] = str[i - 1] == str[j] ? ++j : 0;
            }
            return Arrays.copyOfRange(kmp, 1, n + 1);
        }

        public static int[] findAllOccurrences(String text, String pattern) {
            char[] string = (pattern + "$" + text).toCharArray();
            int[] s = kmpTable(string);
            int p = pattern.length();
            int[] occur = new int[text.length()];
            int q = 0;
            for (int i = p + 1; i < string.length; i++) {
                if (s[i] == p) {
                    occur[q++] = i - 2 * p;
                }
            }
            return Arrays.copyOfRange(occur, 0, q);
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        private int pread() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public String next() {
            return nextString();
        }

        public int nextInt() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = pread();
            }
            int res = 0;
            do {
                if (c == ',') {
                    c = pread();
                }
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = pread();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public String nextString() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = pread();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }
}