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

import sys
import os.path
 
if(os.path.exists('input.txt')) :
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")
    sys.stderr = open("error.txt", "w")
    
depth = 200001
mod = 1000000007 
lim = mod * mod
sys.setrecursionlimit(depth) 
 
linp = lambda: list(minp())
minp = lambda: map(int, input().split())
 
from math import inf, ceil, sqrt, log2, gcd
from collections import defaultdict, deque

dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]

def expo(a, b) :
    if not b : return 1
    v = expo(a, b//2)
    v = (v * v) % mod
    if b%2 : v = (v * a) % mod
    return v

def modinv(a, b) :
    d = expo(b, mod-2)
    v = (a * d) % mod
    return v

def checkPrime(n) :
    if n < 4 : return True
    if n%2 == 0 : return False
    for i in range(3, int(sqrt(n))+1, 2) :
        if n%i == 0 :
            return False
    return True    

def highestPowerof2(x) :
    x |= x >> 1
    x |= x >> 2
    x |= x >> 4
    x |= x >> 8
    x |= x >> 16

    return x ^ (x >> 1)

def isVowel(char):
    return char.lower() in 'aeiou'

for _ in range(int(input())) :
    n, k, x = minp()
    s = str(input())
    a, c = list(), 0
    x = x - 1

    for e in s :
        if e == '*' :
            c = c + 1
        else :
            if c > 0 : a.append(c)
            c = 0
    if c > 0 : a.append(c)

    m = len(a)
    for i in range(m) :
        a[i] = a[i] * k + 1

    b = [a[i] for i in range(m)]
    for i in range(m-2, -1, -1) :
        b[i] = b[i] * b[i+1]
    b.append(1)

    c = list()
    for i in range(m) :
        v = x // b[i+1] 
        c.append(v)
        x = x - v * b[i+1]

    r, f, i = [], True, 0
    for e in s :    
        if e == "*" :
            f = False
        else :
            if not f :
                r.append("b"*c[i])
                i = i + 1
                f = True
            r.append(e)
    if not f : 
        r.append("b"*c[i])

    print("".join(r))