#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;
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

### Response

```cpp
#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;
}

```