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

# cook your dish here
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
sys.setrecursionlimit(3*10**5)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
        
ans=0
a=[]
n,k=0,0
b=[]
vis=[]
tot=0
time=[]
time1=[]
adj=[]
mx=-1
eps=0.0000001
x=0
pref=[]
lvl=[]
gp=[]
ans=[]
f=0
w=[]

def dfs(i,prod,rem,n):
    if rem==0:
        while(i<len(n)):
            prod*=int(n[i])
            i+=1
        return prod
    if i==len(n):
        return prod
    mx=0
    for j in range(9-int(n[i])):
        mx=max(dfs(i+1,(int(n[i])+min(j,rem))*prod,rem-min(j,rem),n),mx)
    return mx
class Disjoint_SU:
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.size=[1 for i in range(n)]
    #find
    def lake_de(self,a):
        if a==self.parent[a]:
            return a
        self.parent[a]=self.lake_de(self.parent[a])
        return self.parent[a]
    #union
    #by size
    def jod(self,a,b):
        x=self.lake_de(a)
        y=self.lake_de(b)
        #for cycle in graph
        if x==y:
            return True
        if self.size[x]>self.size[y]:
            self.size[x]+=self.size[y]
            self.parent[y]=x
        else:
            self.size[y]+=self.size[x]
            self.parent[x]=y
        return False
def hnbhai(tc):
    n,k,x=sd()
    a=sa()
    stck=[]
    tot=0
    are=[]
    for i in range(n):
        if a[i]=="*":
            tot+=1
        else:
            if tot!=0:
                are.append(["b",k*tot+1])
            tot=0
            are.append(["a",1])
    if tot!=0:
        are.append(["b",k*tot+1])
    suffix=[1]*(n+1)
    for i in range(len(are)-1,-1,-1):
        suffix[i]=suffix[i+1]*(are[i][1])
        suffix[i]=min(10**19,suffix[i])
    #x-=1
    ans=[]
    for i in range(len(are)):
        if are[i][0]=="b":
            for j in range(are[i][1]):
                if (j+1)*suffix[i+1]>=x:
                    ans.append("b"*j)
                    #print(x)
                    x-=j*suffix[i+1]
                    break
        else:
            ans.append(are[i][0])
    print("".join(ans))
    return
for _ in range(sb()):
    hnbhai(_+1)
##case

#.
