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.
2
55 982 31372
a*a*aa*aa*a*aa*aa*aaa*aaaaa*a*aa*aaaaa*aaaaa*aa*a*a*a*a
63 9 10946
*aaaaa*aa*a*a*aa*aaaa*a*aa*a*a*aaaa*aaaaa*a*a*aaa*aa*aaa*a*a*aa

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbba
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaaaabbbbbbbbbabbbbabbbbbaa

//package codeforces.edu119;

import java.io.*;
import java.util.*;

import static java.lang.Math.*;

//Think through the entire logic before jump into coding!
//If you are out of ideas, take a guess! It is better than doing nothing!
//Read both C and D, it is possible that D is easier than C for you!

//Be aware of integer overflow!
//If you find an answer and want to return immediately, don't forget to flush before return!

public class C {
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        //initReaderPrinter(true);
        initReaderPrinter(false);
        solve(in.nextInt());
        //solve(1);        
    }

    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            int n = in.nextInt(), k = in.nextInt();
            long x = in.nextLong() - 1;
            char[] s = in.next().toCharArray();
            int bCnt = 0;
            ArrayDeque<int[]> base = new ArrayDeque<>();
            for(int i = 0; i < n; i++) {
                if(s[i] == 'a') {
                    if(bCnt > 0) {
                        base.addLast(new int[]{i - bCnt, bCnt * k + 1});
                    }
                    bCnt = 0;
                }
                else {
                    bCnt++;
                }
            }
            if(bCnt > 0) {
                base.addLast(new int[]{n - bCnt, bCnt * k + 1});
            }
            ArrayDeque<int[]> bq = new ArrayDeque<>();
            while(x > 0) {
                int[] e = base.removeLast();
                bq.addFirst(new int[]{e[0], (int)(x % e[1])});
                x /= e[1];
            }
            StringBuilder ans = new StringBuilder();
            for(int i = 0; i < n; i++) {
                if(s[i] == 'a') ans.append('a');
                else if((i == 0 || s[i] != s[i - 1]) && bq.size() > 0 && bq.peekFirst()[0] == i) {
                    int total = bq.removeFirst()[1];
                    for(int j = 0; j < total; j++) {
                        ans.append('b');
                    }
                }
            }
            out.println(ans);
        }
        out.close();
    }

    static void initReaderPrinter(boolean test) {
        if (test) {
            try {
                in = new InputReader(new FileInputStream("src/input.in"));
                out = new PrintWriter(new FileOutputStream("src/output.out"));
            } catch (IOException e) {
                e.printStackTrace();
            }
        } else {
            in = new InputReader(System.in);
            out = new PrintWriter(System.out);
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        InputReader(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream), 32768);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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;
        }

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitive(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitiveOneIndexed(int n) {
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextInt();
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitive(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitiveOneIndexed(int n) {
            long[] a = new long[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextLong();
            return a;
        }

        String[] nextStringArray(int n) {
            String[] g = new String[n];
            for (int i = 0; i < n; i++) g[i] = next();
            return g;
        }

        List<Integer>[] readGraphOneIndexed(int n, int m) {
            List<Integer>[] adj = new List[n + 1];
            for (int i = 0; i <= n; i++) {
                adj[i] = new ArrayList<>();
            }
            for (int i = 0; i < m; i++) {
                int u = nextInt();
                int v = nextInt();
                adj[u].add(v);
                adj[v].add(u);
            }
            return adj;
        }

        List<Integer>[] readGraphZeroIndexed(int n, int m) {
            List<Integer>[] adj = new List[n];
            for (int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }
            for (int i = 0; i < m; i++) {
                int u = nextInt() - 1;
                int v = nextInt() - 1;
                adj[u].add(v);
                adj[v].add(u);
            }
            return adj;
        }

        /*
            A more efficient way of building a graph using int[] instead of ArrayList to store each node's neighboring nodes.
            1-indexed.
         */
        int[][] buildGraph(int nodeCnt, int edgeCnt) {
            int[] end1 = new int[edgeCnt], end2 = new int[edgeCnt];
            int[] edgeCntForEachNode = new int[nodeCnt + 1], idxForEachNode = new int[nodeCnt + 1];
            for (int i = 0; i < edgeCnt; i++) {
                int u = in.nextInt(), v = in.nextInt();
                edgeCntForEachNode[u]++;
                edgeCntForEachNode[v]++;
                end1[i] = u;
                end2[i] = v;
            }
            int[][] adj = new int[nodeCnt + 1][];
            for (int i = 1; i <= nodeCnt; i++) {
                adj[i] = new int[edgeCntForEachNode[i]];
            }
            for (int i = 0; i < edgeCnt; i++) {
                adj[end1[i]][idxForEachNode[end1[i]]] = end2[i];
                idxForEachNode[end1[i]]++;
                adj[end2[i]][idxForEachNode[end2[i]]] = end1[i];
                idxForEachNode[end2[i]]++;
            }
            return adj;
        }
    }
}