human
PYTHON3 solution for "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...":


 ... What is a Python code to find the x-th lexicographically smallest of them?
assistant
#Here is Python code to find the x-th lexicographically smallest of them.
 def bas(n,k,x,s):
    i=0
    dc=[]
    cd=[]
    st=''
    while(i<len(s)):
        if s[i]=='*':
            j=i+1
            if j<len(s):
                while(s[j]=='*'):
                    j=j+1
                    if j>=len(s):
                        break
            cd.append(j-i)
            i=j-1
        i=i+1
    i=0
    while(i<len(s)):
        if s[i]=='a':
            j=i+1
            if j<len(s):
                while(s[j]=='a'):
                    j=j+1
                    if j>=len(s):
                        break
            dc.append(j-i)
            i=j-1
        i=i+1
    i=0
    while(i<len(cd)):
        cd[i]=cd[i]*k+1
        i=i+1
    i=0
    while(i<len(dc)):
        dc[i]=dc[i]*'a'
        i=i+1
    i=len(cd)-1
    p=1
    while(i>=0):
        if p>=x:
            break
        p=p*cd[i]
        i=i-1
    v=i+1
    i=len(cd)-1
    lans=[0]*len(cd)
    while(i>=v):
        lans[i]=x%cd[i]
        lans[i]=lans[i]-1
        if lans[i]<0:
            lans[i]=lans[i]+cd[i]
        if x%cd[i]!=0:
            x=x//cd[i]+1
        else:
            x=x//cd[i]
        i=i-1
    i=0
    while(i<len(lans)):
        if lans[i]==0:
            lans[i]=''
        else:
            lans[i]=lans[i]*'b'
        i=i+1
    ans=''
    if s[0]=='a':
        i=0
        j=0
        while(i<len(dc)):
            ans=ans+dc[i]
            if j<len(lans):
                ans=ans+lans[j]
                j=j+1
            i=i+1
    else:
        i=0
        j=0
        while(i<len(lans)):
            ans=ans+lans[i]
            if j<len(dc):
                ans=ans+dc[j]
                j=j+1
            i=i+1
    return ans
t=int(input())
l=[]
for x in range(t):
    s1=str(input())
    s2=str(input())
    l.append([s1,s2])
la=[]
for x in l:
    lt=x[0].split(' ',)
    l1=[]
    for y in lt:
        l1.append(int(y))
    s=x[1]
    n=l1[0]
    k=l1[1]
    x=l1[2]
    ans=bas(n,k,x,s)
    la.append(ans)
for x in la:
    print(x)