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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define M 1000000007
#define fio                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
#define d(x) cout << #x << " " << x << "\n";
#define min(x1, x2) (x1 > x2 ? x2 : x1)
#define max(x1, x2) (x1 < x2 ? x2 : x1)
#define min3(x1, x2, x3) (x3 > min(x1, x2) ? min(x2, x1) : x3)
#define max3(x1, x2, x3) (x3 < max(x1, x2) ? max(x1, x2) : x3)
#define ll long long int
#define p pair<ll, ll>
#define ld long double
#define dv(v)                            \
    cerr << #v << "  ";                  \
    for (int i = 0; i < (v).size(); i++) \
        cerr << v[i] << " ";             \
    cerr << "\n";
#define inf INT_MAX
#define mp(x, y) make_pair(x, y)
#define oset tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
using namespace __gnu_pbds;
using namespace std;

int main()
{
    fio;
    ll t, n, x, y, z, k;
    cin >> t;
    while (t--)
    {
        cin >> n >> k >> x;
        string s;
        cin >> s;
        vector<ll> v;
        for (int i = 0; i < n; ++i)
        {
            z = 0;
            while (i < n && s[i] == '*')
            {
                ++z;
                ++i;
            }
            if (z)
            {
                v.push_back(z * k + 1);
                --i;
            }
            else
            {
                v.push_back(0);
            }
        }
        vector<ll> suf;
        z = 1;
        bool b0 = 0;
        for (int i = v.size() - 1; i >= 0; --i)
        {
            if (v[i])
            {
                if (z > 1e18)
                {
                    suf.push_back(-1);
                    continue;
                }
                z *= v[i];
                suf.push_back(z);
            }
        }
        reverse(suf.begin(), suf.end());
        suf.push_back(1);
        suf.push_back(1);
        k = 0;
        for (int i = 0; i < v.size(); ++i)
        {
            if (v[i] == 0)
            {
                cout << "a";
            }
            else
            {
                // suf[k]
                if (suf[k + 1] == -1)
                {
                    z = 0;
                }
                else
                {
                    z = x / suf[k + 1];
                    if (x % suf[k + 1])
                    {
                        ++z;
                    }
                }
                for (int i = 1; i <= z - 1; ++i)
                {
                    cout << "b";
                }
                if (z)
                {
                    x -= (z - 1) * suf[k + 1];
                }
                ++k;
            }
        }
        cout << "\n";
    }
    return 0;
}