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;
  long long int mod = 1000000007;
  constexpr long long inf = 1e18;
  const long long maxn = 2e5 + 100;
  const int MAXN = 1.1e6;
  int spf[MAXN];
  void sieve() {
    for (int i = 1; i < MAXN; i++) spf[i] = 1;
    for (int i = 2; i * i < MAXN; i++) {
      for (int j = i * i; j < MAXN; j += i * i) spf[j] = i;
    }
  }
  long long fact[MAXN], inv[MAXN], invFact[MAXN];
  void factGen() {
    fact[0] = inv[1] = fact[1] = invFact[0] = invFact[1] = 1;
    for (long long i = 2; i < MAXN; i++) {
      fact[i] = (fact[i - 1] * i) % mod;
      inv[i] = mod - (inv[mod % i] * (mod / i) % mod);
      invFact[i] = (inv[i] * invFact[i - 1]) % mod;
    }
  }
  int gcdExtended(int a, int b, int *x, int *y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    int x1, y1;
    int gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  int modInverse(int a, int m) {
    int x, y;
    int g = gcdExtended(a, m, &x, &y);
    if (g != 1)
      return -1;
    else {
      int res = (x % m + m) % m;
      return res;
    }
  }
  long long binpow(long long a, long long b, long long m) {
    if (b < 0) return 0LL;
    if (a <= 0) return 0LL;
    a %= m;
    long long ans = 1LL;
    while (b) {
      if (b & 1) ans = ans * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return ans;
  }
  long long modinv(long long n) { return binpow(n, mod - 2, mod); }
  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 * b) / gcd(a, b); }
  string go(string t) {
    while (t.back() == '0') t.pop_back();
    reverse(t.begin(), t.end());
    return t;
  }
  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 solve() {
    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((int)cnt.size());
    m[(int)m.size() - 1] = 1;
    for (long long i = (int)cnt.size() - 2; i >= 0; --i) {
      m[i] = mul(m[i + 1], k * cnt[i + 1] + 1);
    }
    vector<long long> ans((int)cnt.size());
    for (long long i = 0; i < (int)cnt.size(); ++i) {
      ans[i] = x / m[i];
      x %= m[i];
    }
    for (long long i = 0; i < (int)cnt.size(); ++i) {
      for (long long j = 0; j < ans[i]; ++j) {
        cout << 'b';
      }
      if (i + 1 != (int)cnt.size()) {
        cout << 'a';
      }
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  