Problem: The length of the longest common prefix of two strings s = s_1 s_2 … s_n and t = t_1 t_2 … t_m is defined as the maximum integer k (0 ≤ k ≤ min(n,m)) such that s_1 s_2 … s_k equals t_1 t_2 … t_k.

Koa the Koala initially has n+1 strings s_1, s_2, ..., s_{n+1}.

For each i (1 ≤ i ≤ n) she calculated a_i — the length of the longest common prefix of s_i and s_{i+1}.

Several days later Koa found these numbers, but she couldn't remember the strings.

So Koa would like to find some strings s_1, s_2, ..., s_{n+1} which would have generated numbers a_1, a_2, ..., a_n. Can you help her?

If there are many answers print any. We can show that answer always exists for the given constraints. 

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list a.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 50) — the elements of a.

It is guaranteed that the sum of n over all test cases does not exceed 100.

Output

For each test case:

Output n+1 lines. In the i-th line print string s_i (1 ≤ |s_i| ≤ 200), consisting of lowercase Latin letters. Length of the longest common prefix of strings s_i and s_{i+1} has to be equal to a_i.

If there are many answers print any. We can show that answer always exists for the given constraints.

Example

Input


4
4
1 2 4 2
2
5 3
3
1 3 1
3
0 0 0


Output


aeren
ari
arousal
around
ari
monogon
monogamy
monthly
kevinvu
kuroni
kurioni
korone
anton
loves
adhoc
problems

Note

In the 1-st test case one of the possible answers is s = [aeren, ari, arousal, around, ari].

Lengths of longest common prefixes are:

  * Between \color{red}{a}eren and \color{red}{a}ri → 1 
  * Between \color{red}{ar}i and \color{red}{ar}ousal → 2 
  * Between \color{red}{arou}sal and \color{red}{arou}nd → 4 
  * Between \color{red}{ar}ound and \color{red}{ar}i → 2 

Solution (in PYTHON3):

Solution:
#!/usr/local/bin/python3
"""
Given an array of legnth n, 
you have that arr[i] = the length of the longest common prefix of S[i] and S[i+1]
where S is an array of strings.
Find any possible S.

Observations:
arr = [1,5,4,6,3]
S =
[
    1: 'a','aaaaa'
    5: 'aaaaa', 'aaaaa'
    4: 'aaaaa', 'aaaabb'
    6: 'aaaabb' 'aaaabb',
    3: 'aaaabb', 'aaa'
]

If the seequence is increasing, then you can just keep adding on the same character.
I.E. a local maximum
    If there is something which increases and then ends with something decreasing, then you can just remove letters.
If there is a local minimum, then you need to add a new character. Would just swapping between 'a' and 'b' work? Maybe

arr = [1,5,4,6,5,3,4]
           i 
azzzzz
azzzzz
aaaaaz
aaaazz
aaaazz
aaaaza

____

for i in reverse:
    start with result[-1]
    if arr[i-1] > arr[i]:
        result[-1] += c'*diff
        result.append(result[-1])
    else:
        result.append(result[-1] - diff)

Let's suppose we're at i.
And result[:-1] already meets the criteria.
Now, the only criteria we have to fulfill is that we want the next string to have 
length arr[i] common prefix with result[-1].
    - if len(result[-1]) > len(arr[i]):
        - cut some things off
    else:
        - This is the hard part.

"""

def largestPrefixLength(s1, s2):
    result = 0
    for i in range(min(len(s1),len(s2))):
        if s1[i] == s2[i]:
            result += 1
        else:
            break
    return result

# from lesterlib import lesterutil
# @lesterutil.lesterlog()
def check(arr,result):
    print(result)
    assert(len(result) == len(arr)+1)
    for i in range(len(arr)):
        k = largestPrefixLength(result[i],result[i+1])
        assert(k == arr[i])

import string 
stringMap = {i:c for i,c in enumerate(string.ascii_lowercase)}
def charToNum(c):
    return ord(c)-96

def fast(n,arr):
    result = ['z'*(max(arr)+1)]
    for num in arr:
        curr = list(result[-1][:])
        curr[num] = 'a' if curr[num] == 'z' else 'z'
        result.append(''.join(curr))
    return result

def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        arr = [int(x) for x in input().split()]
        #CHANGE
        result = fast(n,arr)
        for row in result:
            print(row)

        # check(arr,result)

import random
def test():
    arr = [4, 1, 4, 1]
    n = len(arr)
    result = fast(n,arr)
    print(result)
    check(arr,result)

def randomTest():
    while True:
        n = random.randint(1,4)
        arr = [random.randint(0,4) for _ in range(n)]
        result = fast(n,arr)
        print(arr,result)
        check(arr,result)

main()
# test()
# randomTest()