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.IOException;
import java.io.InputStream;
import java.util.*;

public class TestClass {


    static final class InputReader {
        private final InputStream stream;
        private final byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

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

        private int read() throws IOException {
            if (curChar >= numChars) {
                curChar = 0;
                numChars = stream.read(buf);
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public final int readInt() throws IOException {
            return (int) readLong();
        }

        public final long readLong() throws IOException {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
                if (c == -1) throw new IOException();
            }
            boolean negative = false;
            if (c == '-') {
                negative = true;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return negative ? -res : res;
        }

        public final int[] readIntArray(int size) throws IOException {
            int[] array = new int[size];
            for (int i = 0; i < size; i++) {
                array[i] = readInt();
            }
            return array;
        }

        public final long[] readLongArray(int size) throws IOException {
            long[] array = new long[size];
            for (int i = 0; i < size; i++) {
                array[i] = readLong();
            }
            return array;
        }

        public final String readString() throws IOException {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.append((char) c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

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

    static long mulmod(long a, long b,
                       long mod) {
        long res = 0; // Initialize result
        a = a % mod;
        while (b > 0) {
            // If b is odd, add 'a' to result
            if (b % 2 == 1) {
                res = (res + a) % mod;
            }

            // Multiply 'a' with 2
            a = (a * 2) % mod;

            // Divide b by 2
            b /= 2;
        }

        // Return result
        return res % mod;
    }

    static long pow(long a, long b, long MOD) {
        long x = 1, y = a;
        while (b > 0) {
            if (b % 2 == 1) {
                x = (x * y);
                if (x > MOD) x %= MOD;
            }
            y = (y * y);
            if (y > MOD) y %= MOD;
            b /= 2;
        }
        return x;
    }

    static long[] f = new long[100001];

    static long InverseEuler(long n, long MOD) {
        return pow(n, MOD - 2, MOD);
    }

    static long C(int n, int r, long MOD) {

        return (f[n] * ((InverseEuler(f[r], MOD) * InverseEuler(f[n - r], MOD)) % MOD)) % MOD;
    }


    static int[] h = {0, 0, -1, 1};
    static int[] v = {1, -1, 0, 0};


    public static class Pair {
        public int a, b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    static class Pair2 {
        public long cost;
        int node;

        public Pair2(long cos, int node) {
            this.cost = cos;
            this.node = node;
        }
    }

    static long compute_hash(String s) {
        int p = 31;
        int m = 1000000007;
        long hash_value = 0;
        long p_pow = 1;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            hash_value = (hash_value + (c - 'a' + 1) * p_pow) % m;
            p_pow = (p_pow * p) % m;
        }
        return hash_value;
    }

    public static class SegmentTree {
        long[] tree;
        int n;

        public SegmentTree(int[] nodes) {
            tree = new long[nodes.length * 4];
            n = nodes.length;
            build(0, n - 1, 0, nodes);

        }

        private void build(int l, int r, int pos, int[] nodes) {
            if (l == r) {
                tree[pos] = nodes[l];
                return;
            }
            int mid = (l + r) / 2;
            build(l, mid, 2 * pos + 1, nodes);
            build(mid + 1, r, 2 * pos + 2, nodes);
            tree[pos] = tree[2 * pos + 1] + tree[2 * pos + 2];
        }

        public void update(int pos, int val) {
            updateUtil(0, n - 1, 0, pos, val);
        }

        public long get(int l, int r) {
            return getUtil(0, n - 1, 0, l, r);
        }

        private long getUtil(int l, int r, int pos, int ql, int qr) {
            if (ql > r || qr < l) {
                return 0;
            }
            if (l >= ql && r <= qr) {
                return tree[pos];
            }

            int mid = (l + r) / 2;
            return getUtil(l, mid, 2 * pos + 1, ql, qr) + getUtil(mid + 1, r, 2 * pos + 2, ql, qr);

        }

        private void updateUtil(int l, int r, int pos, int i, int val) {
            if (i < l || i > r) {
                return;
            }
            if (l == r) {
                tree[pos] = val;
                return;
            }
            int mid = (l + r) / 2;
            updateUtil(l, mid, 2 * pos + 1, i, val);
            updateUtil(mid + 1, r, 2 * pos + 2, i, val);
            tree[pos] = tree[2 * pos + 1] + tree[2 * pos + 2];
        }
    }

    static int counter = 0;
    static int[] rIn;
    static int[] rOut;
    static int[] lIn;
    static int[] lOut;
    private static int[] flatten;
    private static int[] lFlatten;
    static long answer = 0;

    static int VISITED = 1;
    static int VISITING = 2;

    public static void main(String[] args) throws Exception {
        //https://i...content-available-to-author-only...e.com/ebRGa6
        InputReader in = new InputReader(System.in);


        int t = in.readInt();


        while (t-- > 0) {
            int n = in.readInt();
            String s = in.readString();
            String aba = "abacaba";
            boolean found = false;
            for (int i = 0; i < n; ++i) {
                if (i + 7 <= n) {
                    String newSS = "";
                    for (int j = 0; j < i; ++j) {
                        if (s.charAt(j) == '?') {
                            newSS += 'x';
                        } else {
                            newSS += s.charAt(j);
                        }
                    }
                    String newS = "";
                    for (int j = i; j < i + 7; j++) {
                        if (s.charAt(j) == '?') {
                            newS += aba.charAt(j-i);
                        } else {
                            newS += s.charAt(j);
                        }
                    }
                    if (newS.equals(aba)) {
                        newS = newSS + newS;
                        for (int j = i + 7; j < n; ++j) {
                            if (s.charAt(j) == '?') {
                                newS += 'x';
                            } else {
                                newS += s.charAt(j);
                            }
                        }

//                        System.out.println/(newS);
                        if (countFreq(aba, newS) == 1) {
                            System.out.println("Yes");
                            System.out.println(newS);
                            found = true;
                        }
                    }
                }
                if (found) break;
            }
            if (!found) {
                System.out.println("No");
            }
        }


    }
    static int countFreq(String pattern, String text) {
        int m = pattern.length();
        int n = text.length();
        int res = 0;

        for (int i = 0; i <= n - m; i++) {
            int j;
            for (j = 0; j < m; j++) {
                if (text.charAt(i + j) != pattern.charAt(j)) {
                    break;
                }
            }
            if (j == m) {
                res++;
                j = 0;
            }
        }
        return res;
    }

    private static int solve(int n, int[] dp, int cn) {
        if (cn > n) return 0;
        if (cn == n) return 1;
        if (dp[cn] != -1) return dp[cn];
        int ans = 0;
        for (int i = 2; i <= 9; ++i) {
            ans += solve(n, dp, cn*i);
        }
        return dp[cn] = ans;
    }

    private static void dfs(int node, List<List<Integer>> g, HashSet<Pair> h, ArrayList<Pair> cur, ArrayList<Pair> ans, int[] vis, int last) {
        if (vis[node] == VISITED) return;

        vis[node] = VISITING;
        for (int i = 0; i < g.get(node).size(); ++i) {
            int ntb = g.get(node).get(i);
            if (ntb != last) {
                if (vis[ntb] == VISITING) {
                    if (node > ntb) {

                    }
                } else if (vis[ntb] != VISITED) {
                    if (node > ntb) {
                        if (h.contains(new Pair(ntb, node))) {
                            cur.add(new Pair(ntb, node));
                            dfs(ntb, g, h, cur, ans, vis, node);
                            cur.remove(cur.size()-1);
                        } else {
                            dfs(ntb, g, h, cur, ans, vis, node);
                        }
                    }
                }
            }
        }
        vis[node] = VISITED;
    }



    private static void dfsR(List<List<Integer>> g, int node, int[] v) {
        rIn[node] = counter;
        flatten[counter++] = v[node];

        for (int i = 0; i < g.get(node).size(); ++i) {
            dfsR(g, g.get(node).get(i), v);
        }

        rOut[node] = counter;
        flatten[counter++] = v[node] * -1;
    }

    private static void dfsL(List<List<Integer>> g, int node, int[] v) {
        lIn[node] = counter;
        lFlatten[counter++] = v[node];

        for (int i = 0; i < g.get(node).size(); ++i) {
            dfsL(g, g.get(node).get(i), v);
        }

        lOut[node] = counter;
        lFlatten[counter++] = v[node] * -1;
    }


    private static void preprocess(int pos, int[][] pre, List<List<Integer>> tree, int[] traverse, int depth, int last, int[] tin, int[] tout) {
        tin[pos] = counter++;
        traverse[depth] = pos;

        for (int i = 0; depth - (1 << i) >= 0; ++i) {
            pre[pos][i] = traverse[depth - (1 << i)];
        }

        for (int i = 0; i < tree.get(pos).size(); ++i) {
            if (tree.get(pos).get(i) != last)
                preprocess(tree.get(pos).get(i), pre, tree, traverse, depth + 1, pos, tin, tout);
        }
        tout[pos] = counter++;
    }

    static int gcd(int a, int b) {

        while (b != 0) {
            int t = a;
            a = b;
            b = t % b;
        }
        return a;
    }


    static boolean submit = true;

    static void debug(String s) {
        if (!submit)
            System.out.println(s);
    }

    static void debug(int s) {
        if (!submit)
            System.out.println(s);
    }




}