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

#include <bits/stdc++.h>
using namespace std;
const int maxn = 4 * 1e5 + 10, mod = 1e9 + 7;
const int maxn_mp = 1005;
const int small_inf = 0x3f3f3f3f;
const double EPS = 1e-3;
long long n, m, k, z, h, cnt, flag;
long long ha1 = 227, ha2 = 197;
char c[1005][1005];
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {-1, 0, 1, 0};
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first < b.first;
}
long long f[505][505];
int vis[maxn];
vector<int> G[maxn];
map<string, int> mp;
string s;
long long ans = 0x7fffffff;
bool check() {
  if (1)
    return 1;
  else
    return 0;
}
pair<long long, long long> a[maxn], b[maxn];
void dfs(int i) {
  vis[i] = 1;
  for (int j = 0; j < (int)G[i].size(); j++)
    if (!vis[G[i][j]]) dfs(G[i][j]);
  return;
}
int dp[maxn];
void solve() {
  memset(dp, 0, sizeof(dp));
  int x = 0;
  cin >> n >> k >> x;
  cin >> s;
  int pos = -1;
  dp[n] = 1;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '*' && s[i + 1] == 'a')
      dp[i] = dp[i + 1] * k + dp[i + 1];
    else if (s[i] == '*')
      dp[i] = dp[i + 1] + k;
    else
      dp[i] = dp[i + 1];
  }
  for (int i = 0; i < n; i++) {
    cout << dp[i] << ' ';
    if (dp[i] >= x) pos = i;
  }
  cout << pos;
  cout << endl;
  for (int i = 0; i < pos; i++)
    if (s[i] == 'a') cout << 'a';
  for (int i = pos; i < n; i++) {
    if (s[i] == 'a') continue;
    while (x > dp[i + 1]) {
      cout << 'b';
      x -= dp[i + 1];
    }
  }
  cout << endl;
}
signed main() {
  int ttt = 0;
  int tt;
  cin >> tt;
  while (tt--) solve();
  ios::sync_with_stdio(false);
  cin.tie(0);
  return 0;
}
