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 PYTHON):

Solution:


# template by 3xC and starkizard.
# contributors: 
    
#####################################################################################
from __future__ import division, print_function 
    
import sys
import os
from collections import Counter, deque, defaultdict
import itertools
import math
import io
    
    
"""Uncomment modules according to your need"""
    
    
# from bisect import bisect_left, bisect_right, insort
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from copy import deepcopy
# from decimal import *
# import re
# import operator
    
#####################################################################################
# this enables you to write python3 code with PyPy2 (Python 2)
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange
    
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
    
#####################################################################################
    
"""value of mod"""
MOD = 998244353
mod = 10**9 + 7
    
"""Uncomment next 4 lines if doing huge recursion"""
# import threading
# threading.stack_size(1<<27)
# sys.setrecursionlimit(10000
    
def prepare_factorial(mod=mod):
    """ returns two lists, factorial and inverse factorial modulo argument by default 10**9 +7 """
    # Comment code out when you don't need inverse factorial or vice versa
    fact = [1]
    for i in range(1, 200005):
        fact.append((fact[-1] * i) % mod)
    
    ifact = [0] * 200005
    ifact[200004] = pow(fact[200004], mod - 2, mod)
    for i in range(200004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod
    
    return fact, ifact
    
def modinv(n, p):
    """ returns N inverse modulo p """
    return pow(n, p - 2, p)
    
def ncr(n, r,  fact, ifact):
    """ takes 4 arguments: n , r and factorial and inverse factorial lists"""
    t = (fact[n] * (ifact[r]*ifact[n-r]) % MOD)% MOD
    return t
    
    
def get_n(Sum):    
    """this function returns the maximum n for which Summation(n) <= Sum"""
    ans = (-1 + sqrt(1 + 8*Sum))//2
    return ans
    
def sieve(n):
    """ returns a list of prime numbers till n """
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
    
    
    
def divs(n, start=1):
    """ returns a list of all divisors till n """
    divisors = []
    #rdivisors=[]
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors
    
    
    
def divn(N):
    """ returns the number of divisors """
    divs_number = 1
    factorCounter=1
    tempList=[]
    
    while N%2==0:
        factorCounter+=1
        N=N//2
    tempList.append(factorCounter)
    
    i=3
    
    while i*i<=N:
        factorCounter=1
        while N%i==0:
            factorCounter+=1
            N=N//i
        if factorCounter!=1:
            tempList.append(factorCounter)
        i+=1
    
    if N>2:
        factorCounter=2
        tempList.append(factorCounter)
    
    divisorCount=1
    for i in tempList:
        divisorCount*=i
    return(divisorCount)
    
    
    
def lrfind(d, x, default=-1):
    """ Takes 2 arguments an iterable and an element. returns a tuple (firstoccurence,lastoccurence) -1 if not found """
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right
    
def gcd(x, y): # math.gcd is slower
    """ returns greatest common divisor of x and y """
    while y:
        x, y = y, x % y
    return x
    
def check_sorted(a):
    ''' returns True/False '''
    for i in range(len(a)-1):
        if a[i]>a[i+1]:
            return False
    return True
def ceil(n, k=1): return n // k + (n % k != 0) #returns math.ceil but protecting against floating inconsistencies
def input(): return sys.stdin.readline().strip()
def ii(): return int(input()) #inputs integer
def mi(): return map(int, input().split()) # inputting space seperated variables for example x,y,z
def li(): return list(map(int, input().split())) #inputting a space seperated list of integers
def lw(): return input().split() #inputting a space seperated list of strings
def lcm(a, b): return abs(a * b) // gcd(a, b) #returns LCM of two arguments
def prr(a, sep=' ', end='\n'): print(sep.join(map(str, a)), end=end) #For printing an iterable with seperator sep as optional second argument (default : " "), ending character (default: "\n") as optional third
def dd(): return defaultdict(int) #returns a dictionary with values defaulted to 0
def ddl(): return defaultdict(list) #returns a dictionary with values defaulted to []
def write(s): return sys.stdout.write(s)
    
    
    
    
    
    
    
    
    
    
    
    
###################################################################
    
def main():
    #CODE GOES HERE:
    for t in range(ii()):
        n=ii()
        a=li()
        s=""
        maxl=max(a)
        if maxl==0:
            s="a"
            maxl=1
        else:
            for i in range(maxl):
                s+=chr(97+(i%26))
        print(s)
        for i in range(n):
            t=""
            for j in range(maxl):
                if j<a[i]:
                    t+=s[j]
                else:
                    c=(ord(s[j])+1)
                    if c>ord('z'):
                        c-=26
                    t+=chr(c)
            s=t
            print(s)
    

        
            
    
    
""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""
    
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
    
import os, sys
from io import IOBase, BytesIO
    
BUFSIZE = 8192
    
class FastIO(BytesIO):
    newlines = 0
    
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
    
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s
    
    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()
    
    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
    
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
    
    
class IOWrapper(IOBase):
    
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')
    
    
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
    
""" main function"""
    
if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()