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: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  
  public class P1 {
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		StringBuilder st=new StringBuilder();
  		while(t--!=0){
  			int n=sc.nextInt();
  			int k=sc.nextInt();
  			long x=sc.nextLong();
  			String s=sc.nextLine();
  			ArrayDeque<Character> ans=new ArrayDeque<>();
  			int count=0;
  			ArrayList<Integer> maintain=new ArrayList<Integer>();
  			for(int i=0;i<n;i++) {
  				if(s.charAt(i)=='*') {
  					count+=k;
  				}else {
  					if(count>0) maintain.add(count);
  					count=0;
  				}
  			}
  			if(count>0) maintain.add(count);
  			long suffix[]=new long[maintain.size()];
  			for(int i=0;i<maintain.size();i++) {
  				suffix[i]=maintain.get(i)+1;
  			}
  			for(int i=maintain.size()-2;i>=0;i--) {
  				if (1000000000000000000l / suffix[i + 1] < suffix[i])
  					suffix[i] = 1000000000000000000l;
  				else
  					suffix[i]=suffix[i]*suffix[i+1];
  			}
  //			for(int i=0;i<maintain.size();i++) {
  //				System.out.print(suffix[i]+" ");
  //			}
  			int current=0;
  			for(int i=0;i<n;i++) {
  				if(s.charAt(i)=='a') {
  					ans.addLast('a');
  				}else if(x>0){
  					while(i<n&&s.charAt(i)=='*') i++;
  					if(i<n) i--;
  					current++;
  					if(current>=maintain.size()) {
  						while(x>1) {
  							x--;
  							ans.addLast('b');
  						}
  					}else {
  						if(suffix[current]>x) {
  							continue;
  						}else {
  							int counta=1;
  							while(counta*suffix[current]<x) {
  								counta++;
  							}
  							x-=((counta-1)*suffix[current]);
  							while(counta>1) {
  								counta--;
  								ans.addLast('b');
  							}
  						}
  					}
  				}
  			}
  			for(char ch:ans) {
  				st.append(ch);
  			}
  			st.append("\n");
  		}
  		System.out.println(st);
  	}
  	
  	static class DisjointUnionSets {
  		int[] rank, parent;
  		int n;
  		
  		// Constructor
  		public DisjointUnionSets(int n)
  		{
  			rank = new int[n];
  			parent = new int[n];
  			this.n = n;
  			makeSet();
  		}
  		
  		// Creates n sets with single item in each
  		void makeSet()
  		{
  			for (int i = 0; i < n; i++) {
  				// Initially, all elements are in
  				// their own set.
  				parent[i] = i;
  			}
  		}
  		
  		// Returns representative of x's set
  		int find(int x)
  		{
  			// Finds the representative of the set
  			// that x is an element of
  			if (parent[x] != x) {
  				// if x is not the parent of itself
  				// Then x is not the representative of
  				// his set,
  				parent[x] = find(parent[x]);
  				
  				// so we recursively call Find on its parent
  				// and move i's node directly under the
  				// representative of this set
  			}
  			
  			return parent[x];
  		}
  		
  		// Unites the set that includes x and the set
  		// that includes x
  		void union(int x, int y)
  		{
  			// Find representatives of two sets
  			int xRoot = find(x), yRoot = find(y);
  			
  			// Elements are in the same set, no need
  			// to unite anything.
  			if (xRoot == yRoot)
  				return;
  			
  			// If x's rank is less than y's rank
  			if (rank[xRoot] < rank[yRoot])
  				
  				// Then move x under y  so that depth
  				// of tree remains less
  				parent[xRoot] = yRoot;
  			
  			// Else if y's rank is less than x's rank
  			else if (rank[yRoot] < rank[xRoot])
  				
  				// Then move y under x so that depth of
  				// tree remains less
  				parent[yRoot] = xRoot;
  			
  			else // if ranks are the same
  			{
  				// Then move y under x (doesn't matter
  				// which one goes where)
  				parent[yRoot] = xRoot;
  				
  				// And increment the result tree's
  				// rank by 1
  				rank[xRoot] = rank[xRoot] + 1;
  			}
  		}
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  	static void sort(char[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			char tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  }
  