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: /*
  Codeforces
  Problem 1620C
  */
  
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.util.ArrayDeque;
  import java.util.Deque;
  
  public class BAString {
  	
  	public static void main(String[] args) throws IOException {
  		FastIO in = new FastIO(System.in);
  		
  		int t = in.nextInt();
  		while (t-- > 0) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			long x = in.nextLong() - 1;
  			StringBuilder s = new StringBuilder(in.next());
  			Deque<Integer> l = new ArrayDeque<>();
  			
  			int cnt = 0;
  			for (int i = 0; i < s.length(); i++) {
  				if (s.charAt(i) == '*') {
  					cnt++;
  				} else if (cnt != 0) {
  					l.addLast(cnt);
  					cnt = 0;
  				}
  			}
  			l.addLast(cnt);
  			
  			int[] dig = new int[l.size()];
  			for (int i = l.size() - 1; i >= 0; i--) {
  				int next = k * l.removeLast() + 1;
  				dig[i] = (int) (x % next);
  				x /= next;
  			}
  			
  			for (int i = 1; i < s.length(); i++) {
  				if (s.charAt(i) == '*' && s.charAt(i-1) == '*') s.deleteCharAt(i--);
  			}
  			
  			int j = 0;
  			for (int i = 0; i < s.length(); i++) {
  				if (s.charAt(i) == '*') {
  					s.replace(i, i+1, str(dig[j]));
  					i = i + dig[j++] - 1;
  				}
  			}
  			
  			System.out.println(s);
  		}
  		
  		in.close();
  	}
  	
  	public static String str(int i) {
  		StringBuilder sb = new StringBuilder();
  		for (int j = 0; j < i; j++) sb.append('b');
  		return sb.toString();
  	}
  	
  	public static class FastIO {
  		private InputStream dis;
  		private byte[] buffer = new byte[1 << 17];
  		private int pointer = 0;
  	
  		public FastIO(String fileName) throws IOException {
  			dis = new FileInputStream(fileName);
  		}
  	
  		public FastIO(InputStream is) throws IOException {
  			dis = is;
  		}
  	
  		public int nextInt() throws IOException {
  			int ret = 0;
  			byte b;
  			
  			do {
  				b = nextByte();
  			} while (b <= ' ');
  			
  			boolean negative = false;
  			if (b == '-') {
  				negative = true;
  				b = nextByte();
  			}
  			
  			while (b >= '0' && b <= '9') {
  				ret = 10 * ret + b - '0';
  				b = nextByte();
  			}
  	
  			return (negative) ? -ret : ret;
  		}
  	
  		public long nextLong() throws IOException {
  			long ret = 0;
  			byte b;
  			
  			do {
  				b = nextByte();
  			} while (b <= ' ');
  			
  			boolean negative = false;
  			if (b == '-') {
  				negative = true;
  				b = nextByte();
  			}
  			
  			while (b >= '0' && b <= '9') {
  				ret = 10 * ret + b - '0';
  				b = nextByte();
  			}
  			
  			return (negative) ? -ret : ret;
  		}
  	
  		public byte nextByte() throws IOException {
  			if (pointer == buffer.length) {
  				dis.read(buffer, 0, buffer.length);
  				pointer = 0;
  			}
  			return buffer[pointer++];
  		}
  	
  		public String next() throws IOException {
  			StringBuffer ret = new StringBuffer();
  			byte b;
  			
  			do {
  				b = nextByte();
  			} while (b <= ' ');
  			
  			while (b > ' ') {
  				ret.appendCodePoint(b);
  				b = nextByte();
  			}
  	
  			return ret.toString();
  		}
  	
  		public void close() throws IOException {
  			dis.close();
  		}
  	}
  }