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: #include <bits/stdc++.h>
  using namespace std;
  #pragma GCC diagnostic ignored "-Wunused-variable"
  #pragma GCC diagnostic ignored "-Wunknown-pragmas"
  #pragma GCC optimize("Ofast")
  #pragma GCC target( \
      "fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
  #pragma GCC optimize("unroll-loops")
  constexpr long long SZ = 1e5 + 7;
  constexpr long long inf = 1e18;
  constexpr long long mod = 1e9 + 7;
  constexpr long long MOD = 998244353;
  constexpr long double PI = 3.141592653589793238462;
  void __print(long long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << x; }
  void __print(const char *x) { cerr << x; }
  void __print(const string &x) { cerr << x; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ", ";
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    long long f = 0;
    cerr << '{';
    for (auto &i : x) {
      cerr << (f++ ? ", " : "");
      __print(i);
    }
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) {
      cerr << ", ";
    }
    _print(v...);
  }
  template <class T>
  bool amin(T &a, const T &b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool amax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &istream, pair<T1, T2> &p) {
    return (istream >> p.first >> p.second);
  }
  template <typename T>
  istream &operator>>(istream &istream, vector<T> &v) {
    for (auto &it : v) {
      cin >> it;
    }
    return istream;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
    return (ostream << p.first << " " << p.second);
  }
  template <typename T>
  ostream &operator<<(ostream &ostream, const vector<T> &c) {
    for (auto &it : c) {
      cout << it << " ";
    }
    return ostream;
  }
  long long tc_cnt = 1;
  long long modpow(long long a, long long b, long long m = mod) {
    a = a & m;
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans = (ans * a) % m;
      }
      b = b >> 1;
      a = (a * a) % m;
    }
    return ans;
  }
  long long modinv(long long a, long long m = mod) { return modpow(a, m - 2); }
  long long modadd(long long a, long long b, long long m = mod) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long modsub(long long a, long long b, long long m = mod) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long modmul(long long a, long long b, long long m = mod) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long moddiv(long long a, long long b, long long m = mod) {
    a = a % m;
    b = b % m;
    return (modmul(a, modinv(b, m), m) + m) % m;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  const long long dx[8] = {0, 1, -1, 0, -1, -1, 1, 1};
  const long long dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  long long mul(long long a, long long b) {
    long long ans = 0;
    while (b) {
      if (b & 1) {
        ans += a;
      }
      b = b >> 1;
      a += a;
      if (a > inf) {
        a = inf;
      }
      if (ans > inf) {
        ans = inf;
      }
    }
    return ans;
  }
  void oreo() {
    long long n, k, x;
    cin >> n >> k >> x;
    x--;
    string s;
    cin >> s;
    vector<long long> cnt;
    long long c = 0;
    for (long long i = 0; i < n; ++i) {
      if (s[i] == 'a') {
        cnt.push_back(c);
        c = 0;
      } else {
        c++;
      }
    }
    cnt.push_back(c);
    ;
    vector<long long> m((long long)cnt.size());
    m[(long long)m.size() - 1] = 1;
    for (long long i = (long long)cnt.size() - 2; i >= 0; --i) {
      m[i] = mul(m[i + 1], k * cnt[i + 1] + 1);
    };
    vector<long long> ans((long long)cnt.size());
    for (long long i = 0; i < (long long)cnt.size(); ++i) {
      ans[i] = x / m[i];
      x %= m[i];
    };
    for (long long i = 0; i < (long long)cnt.size(); ++i) {
      for (long long j = 0; j < ans[i]; ++j) {
        cout << 'b';
      }
      if (i + 1 != (long long)cnt.size()) {
        cout << 'a';
      }
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) oreo();
    return 0;
  }
  