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.*;
import java.math.BigInteger;
import java.util.*;


public class Equal {
    static class pair implements Comparable<pair> {
        int v, w;

        public pair(int v, int w) {
            this.v = v;
            this.w = w;
        }

        @Override
        public int compareTo(pair o) {
            if (v == o.v)
                return w - o.w;
            return o.v - v;
        }
    }

    static class tri implements Comparable<tri> {
        int v, i, j;

        public tri(int v, int i, int j) {
            this.v = v;
            this.i = i;
            this.j = j;
        }

        @Override
        public int compareTo(tri o) {

            return v - o.v;
        }
    }

    static ArrayList<Integer>[] tree;
    static int timer = 0;
    static int[] timein;
    static int[] timeout;
    static int[][] powerParent;
    static int[] depth;
    static int[][] minWeight;

    static void dfs(int u, int parent) {

        timein[u] = ++timer;
//        powerParent[u][0] = parent;
////        minWeight[u][0] = weight;
//        if (parent == -1)
//            powerParent[u][0] = u;
//        for (int i = 1; i <= 32; i++) {
//            powerParent[u][i] = powerParent[powerParent[u][i - 1]][i - 1];
//            minWeight[u][i] = Math.min(minWeight[u][i - 1], minWeight[powerParent[u][i - 1]][i - 1]);
//        }
        for (int i = 0; i < tree[u].size(); i++) {
            int v = tree[u].get(i);
            if (v == parent)
                continue;
            depth[v] = depth[u] + 1;
            dfs(v, u);
        }


        timeout[u] = timer;
    }

    static boolean isAncestor(int u, int v) {
        return timein[u] <= timein[v] && timeout[v] <= timeout[u];
    }

    static int lca(int u, int v) {
        if (isAncestor(u, v))
            return u;
        if (isAncestor(v, u))
            return v;
        for (int i = 32; i >= 0; i--) {
            int temp = powerParent[u][i];
            if (!isAncestor(temp, v)) {
                u = temp;
            }
        }
        return powerParent[u][0];
    }

    static int distance(int u, int v) {
        return depth[u] + depth[v] - 2 * depth[lca(u, v)];
    }

    static int nthAncestor(int u, int n) {
        int c = -1;

        while (n > 0) {
            c++;
            if (n % 2 == 0) {
                n /= 2;
                continue;
            }
            u = powerParent[u][c];
            n /= 2;

        }
        return u;
    }

    static int nthMinWeight(int u, int n) {
        int c = -1;

        int min = (int) (1e9);
        while (n > 0) {
            c++;
            if (n % 2 == 0) {
                n /= 2;
                continue;
            }
            min = Math.min(min, minWeight[u][c]);
            u = powerParent[u][c];
            n /= 2;

        }
        return min;
    }


    static class SegmentTree {        // 1-based DS, OOP

        int N;            //the number of elements in the array as a power of 2 (i.e. after padding)
        int[] array, lazy;
        double[] sTree;

        public SegmentTree(int[] in) {
            array = in;
            N = in.length - 1;
            sTree = new double[N << 1];        //no. of nodes = 2*N - 1, we add one to cross out index zero
            lazy = new int[N << 1];
            build(1, 1, N);
        }

        public void build(int node, int b, int e)    // O(n)
        {
            if (b == e)
                sTree[node] = array[b];
            else {
                int mid = b + e >> 1;
                build(node << 1, b, mid);
                build(node << 1 | 1, mid + 1, e);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
            }
        }


        public void update_point(int index, double oval, double nval)            // O(log n)
        {
            index += N - 1;
            sTree[index] -= oval;
            sTree[index] += nval;
            while (index > 1) {
                index >>= 1;
                sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
            }
        }


        public void update_range(int i, int j, double oval, double nval)        // O(log n)
        {
            update_range(1, 1, N, i, j, oval, nval);
        }

        public void update_range(int node, int b, int e, int i, int j, double oval, double nval) {
            if (i > e || j < b)
                return;
            if (b >= i && e <= j) {
                sTree[node] -= oval;
                sTree[node] += nval;

            } else {
                int mid = b + e >> 1;
//                propagate(node, b, mid, e);
                update_range(node << 1, b, mid, i, j, oval, nval);
                update_range(node << 1 | 1, mid + 1, e, i, j, oval, nval);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
            }

        }

//        public void propagate(int node, int b, int mid, int e) {
//            lazy[node << 1] += lazy[node];
//            lazy[node << 1 | 1] += lazy[node];
//            sTree[node << 1] += (mid - b + 1) * lazy[node];
//            sTree[node << 1 | 1] += (e - mid) * lazy[node];
//            lazy[node] = 0;
//        }

        public double query(int i, int j) {
            return query(1, 1, N, i, j);
        }

        public double query(int node, int b, int e, int i, int j)    // O(log n)
        {
            if (i > e || j < b)
                return 0;
            if (b >= i && e <= j)
                return sTree[node];
            int mid = b + e >> 1;
//            propagate(node, b, mid, e);
            double q1 = query(node << 1, b, mid, i, j);
            double q2 = query(node << 1 | 1, mid + 1, e, i, j);
            return q1 + q2;

        }
    }

    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    static int NumOfOccIn(char[]c){
        int ans=0;
        char[] pattern={'a','b','a','c','a','b','a'};
        for (int i = 0; i <c.length ; i++) {
            int j=i;
            for (; j <c.length ; j++) {
                if(j==i+7)
                    break;
                if(c[j]!=pattern[j-i])
                    break;
            }
            if(j==i+7)
                ans++;
        }
        return ans;
    }
    static char[]cc;
    static int NumOccAfterUpdate(char[]c,int start){
        int ans=0;
       cc=c.clone();
        char[] pattern={'a','b','a','c','a','b','a'};
        for (int i = start; i <c.length ; i++) {
            if(i==start+7)
                break;
            if(cc[i]=='?'){
                cc[i]=pattern[i-start];
            }
            if(cc[i]==pattern[i-start])
                continue;
            else return 0;
        }
        return NumOfOccIn(cc);
    }

    public static void main(String[] args) throws IOException {
        //BufferedReader br = new BufferedReader(new FileReader("name.in"));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter out = new PrintWriter(System.out);
//       char[]c={'h'};
//       NumOccAfterUpdate(c,0);
//       out.println(c[0]);
        int T=Integer.parseInt(br.readLine());
        while (T-->0){
            int n=Integer.parseInt(br.readLine());
            char[]c=br.readLine().toCharArray();
            int current=NumOfOccIn(c);
            if(current>1){
                out.println("NO");
            }
            else {
                if(current==1){
                    out.println("YES");
                    for (int i = 0; i <c.length ; i++) {
                        if(c[i]=='?')
                            c[i]='x';
                        out.print(c[i]);

                    }
                    out.println();
                }
                else {
                    boolean found=false;
                    for (int i = 0; i <c.length ; i++) {
                        if(NumOccAfterUpdate(c,i)==1){
                            found=true;
                            out.println("YES");
                            for (int j = 0; j <cc.length ; j++) {
                                if (cc[j] == '?')
                                    cc[j] = 'x';

                                out.print(cc[j]);

                            }
                            out.println();break;

                        }
                    }
                    if(found)
                        continue;
                    else out.println("NO");

                }
            }

        }
        out.flush();
        out.close();

    }


}