description: You are given an integer k and a string s that consists only of characters 'a' (a lowercase Latin letter) and '*' (an asterisk).
  
  Each asterisk should be replaced with several (from 0 to k inclusive) lowercase Latin letters 'b'. Different asterisk can be replaced with different counts of letter 'b'.
  
  The result of the replacement is called a BA-string.
  
  Two strings a and b are different if they either have different lengths or there exists such a position i that a_i ≠ b_i.
  
  A string a is lexicographically smaller than a string b if and only if one of the following holds: 
  
    * a is a prefix of b, but a ≠ b; 
    * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b. 
  
  
  
  Now consider all different BA-strings and find the x-th lexicographically smallest of them.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.
  
  The first line of each testcase contains three integers n, k and x (1 ≤ n ≤ 2000; 0 ≤ k ≤ 2000; 1 ≤ x ≤ 10^{18}). n is the length of string s.
  
  The second line of each testcase is a string s. It consists of n characters, each of them is either 'a' (a lowercase Latin letter) or '*' (an asterisk).
  
  The sum of n over all testcases doesn't exceed 2000. For each testcase x doesn't exceed the total number of different BA-strings. String s contains at least one character 'a'.
  
  Output
  
  For each testcase, print a single string, consisting only of characters 'b' and 'a' (lowercase Latin letters) — the x-th lexicographically smallest BA-string.
  
  Example
  
  Input
  
  
  3
  2 4 3
  a*
  4 1 3
  a**a
  6 3 20
  **a***
  
  
  Output
  
  
  abb
  abba
  babbbbbbbbb
  
  Note
  
  In the first testcase of the example, BA-strings ordered lexicographically are: 
  
    1. a
    2. ab
    3. abb
    4. abbb
    5. abbbb
  
  
  
  In the second testcase of the example, BA-strings ordered lexicographically are: 
  
    1. aa
    2. aba
    3. abba
  
  
  
  Note that string "aba" is only counted once, even though there are two ways to replace asterisks with characters 'b' to get it.
solution: import java.util.*;
  
  import java.io.*;
  
  public class tr0 {
  	static PrintWriter out;
  	static StringBuilder sb;
  	static long mod = (long) 998244353;
  	static long inf = (long) 1e16;
  	static int n, m;
  	static ArrayList<Integer>[] ad;
  	static ArrayList<int[]>[] quer;
  	static int[][] remove, add;
  	static int[][][] memo;
  	static boolean vis[];
  	static long[] inv, f, ncr[];
  	static HashMap<Integer, Integer> hm;
  	static int[] pre, suf, Smax[], Smin[];
  	static int idmax, idmin;
  	static ArrayList<Integer> av;
  	static HashMap<Integer, Integer> mm;
  	static boolean[] msks;
  	static int[] lazy[], lazyCount;
  	static int[] dist;
  	static int[][] P;
  	static long N;
  	static ArrayList<Integer> gl;
  	static int[] a, c, ans;
  	static String s;
  
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		out = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			long x = sc.nextLong() - 1;
  			String s = sc.nextLine();
  			sb = new StringBuilder();
  			ArrayList<Long> art = new ArrayList<>();
  			for (int i = 0; i < n; i++) {
  				int len = 0;
  				while (i < n && s.charAt(i) == '*') {
  					i++;
  					len++;
  				}
  				if (len != 0) {
  					long num = len * 1l * k;
  					art.add(++num);
  				}
  			}
  			int id = 0;
  			lop: for (int i = 0; i < n; i++) {
  				if (s.charAt(i) == 'a')
  					sb.append('a');
  				else {
  					while (i < n && s.charAt(i) == '*') {
  						i++;
  					}
  					i--;
  					long val = 1;
  					id++;
  					for (int j = art.size() - 1; j >= id; j--) {
  
  						val = val * art.get(j);
  						if (val > x || val < 0)
  							continue lop;
  					}
  					long bs = x / val;
  					x = x % val;
  					while (bs-- > 0) {
  						sb.append('b');
  					}
  				}
  			}
  			out.println(sb.toString());
  		}
  		out.flush();
  	}
  
  //a*a*a
  // aaa
  // aaba
  // aabba
  // abaa
  // ababa
  // ababba
  // abbaa
  // abbaba
  // abbabba
  	static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public Long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextArrInt(int n) throws IOException {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		public long[] nextArrLong(int n) throws IOException {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		public int[] nextArrIntSorted(int n) throws IOException {
  			int[] a = new int[n];
  			Integer[] a1 = new Integer[n];
  			for (int i = 0; i < n; i++)
  				a1[i] = nextInt();
  			Arrays.sort(a1);
  			for (int i = 0; i < n; i++)
  				a[i] = a1[i].intValue();
  			return a;
  		}
  
  		public long[] nextArrLongSorted(int n) throws IOException {
  			long[] a = new long[n];
  			Long[] a1 = new Long[n];
  			for (int i = 0; i < n; i++)
  				a1[i] = nextLong();
  			Arrays.sort(a1);
  			for (int i = 0; i < n; i++)
  				a[i] = a1[i].longValue();
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }