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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  /*
  1
  4 1 3
  a**a
  
  1
  200 1 1000000000000000000
  a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*
  
  1
  3 0 1
  aaa
   */
  public class C {
  
  	static long oo=(long)2e18;
  	static long[][] dp;
  	static int n, k;
  	static long x;
  	static char[] line;
  	static int[] nextA;
  	static PrintWriter out;
  	
  	static boolean killMePls = false;
  	public static void main(String[] args) throws Exception {
  		Thread.UncaughtExceptionHandler h = new Thread.UncaughtExceptionHandler() {
  			public void uncaughtException(Thread t, Throwable e) {killMePls = true;}
  		};
  		Thread t = new Thread(null, ()->A(args), "", 1<<27);
  		t.setUncaughtExceptionHandler(h);
  		t.start();
  		t.join();
  		if(killMePls) throw null;
  	}
  	public static void A(String args[]) {
  		FastScanner fs=new FastScanner();
  		out=new PrintWriter(System.out);
  		int T=fs.nextInt();
  		for (int tt=0; tt<T; tt++) {
  			n=fs.nextInt(); k=fs.nextInt(); x=fs.nextLong();
  			line=fs.next().toCharArray();
  			dp=new long[n][k+1];
  			for (int i=0; i<dp.length; i++) Arrays.fill(dp[i], -1);
  			nextA=new int[n];
  			int nextAInd=n;
  			for (int i=n-1; i>=0; i--) {
  				nextA[i]=nextAInd;
  				if (line[i]=='a') nextAInd=i;
  			}
  			
  			for (int originalPos=n-1; originalPos>=0; originalPos--) {
  				for (int kk=k; kk>=0; kk--) {
  					go(originalPos, kk);
  				}
  			}
  			//TODO: build it
  			x--;
  			print(x, 0, 0);
  			out.println();
  		}
  		out.close();
  	}
  	
  	static long go(int originalPos, int nBs) {
  		if (originalPos==n) return 1;
  		if (dp[originalPos][nBs]!=-1) return dp[originalPos][nBs];
  		//place b or not
  		long ans=0;
  		{
  			//not place b
  			if (line[originalPos]!='*') {
  				ans+=go(originalPos+1, 0);
  			}
  			else if (nBs==k) {
  				ans+=go(originalPos+1, 0);
  			}
  			else {
  				ans+=go(nextA[originalPos], 0);
  			}
  		}
  		{
  			//place b
  			if (line[originalPos]=='*' && nBs<k) {
  				ans+=go(originalPos, nBs+1);
  			}
  		}
  		
  		return dp[originalPos][nBs]=Math.min(ans, oo);
  	}
  	
  	static void print(long x, int originalPos, int nBs) {
  		while (true) {
  			if (originalPos==n)
  				return;
  			if (line[originalPos]=='a') {
  				out.print('a');
  				originalPos++;
  				nBs=0;
  				continue;
  			}
  			//cost of not printing a b
  			int nextInd=nBs<k?nextA[originalPos]:originalPos+1;
  			long costNow=go(nextInd, 0);
  			if (costNow>x) {
  				if (nBs<k) {
  					originalPos=nextA[originalPos];
  					nBs=0;
  				}
  				else {
  					originalPos++;
  					nBs=0;
  				}
  			}
  			else {
  				out.print("b");
  				x-=costNow;
  				nBs++;
  			}
  		}
  	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  
  	
  }
  