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

sr=lambda: input()
ir=lambda: int(sr())
lr=lambda: list(map(int, sr().split()))

inf = 10**18
# mod = 10**9+7
mod = 998244353

if __name__=='__main__':
    test = ir()
    for t in range(test):
        n,k,x = lr()
        a=sr()
        count = []
        pre = ''
        now = 0
        start_inds = []
        for c in a:
            if c == '*':
                if pre == '*':
                    now+=1
                else:
                    now=1
            else:
                if now > 0:
                    count.append(now*k+1)
                now = 0
            pre = c
        if now>0 and pre == '*':
            count.append(now*k+1)
        count.reverse()
        count_per = [1]
        x-=1
        start = 0
        for ind, num in enumerate(count[:-1]):
            count_per.append(num*count_per[-1])
            if count_per[-1] > 10:
                start = len(count)-1-ind
                count_per.pop()
                break
        for i in range(start):
            count.pop()

        count.reverse()
        count_per.reverse()

        nums = []
        ret = x
        for num in count_per:
            tmp = ret//num
            nums.append(tmp)
            ret = ret%num

        ans = []
        now_count = 0
        now = 0
        for c in a:
            if c=='*':
                if pre=='*':
                    now+=1
                else:
                    now=1
            else:
                if now > 0:
                    if now_count >= start:
                        ind = now_count-start
                        for _ in range(nums[ind]):
                            ans.append('b')
                    now_count+=1
                ans.append('a')
                now = 0
            pre=c
        if pre=='*':
            if now_count >= start:
                ind = now_count-start
                for _ in range(nums[ind]):
                    ans.append('b')
            now_count+=1
        print(*ans, sep="")