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:
import bisect
import os
import gc
import sys
from io import BytesIO, IOBase
from collections import Counter
from collections import deque
import heapq
import math
import statistics
def sin():
    return input()
def ain():
    return list(map(int, sin().split()))
def sain():
    return input().split()
def iin():
    return int(sin())
MAX = float('inf')
MIN = float('-inf')
MOD = 1000000007
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    s = set()
    for p in range(2, n+1): 
        if prime[p]: 
            s.add(p)
    return s
def readTree(n, m):
    adj = [deque([]) for _ in range(n+1)]
    for _ in range(m):
        u,v = ain()
        adj[u].append(v)
        adj[v].append(u)
    return adj


def main():
    for _ in range(iin()):
        n = iin()
        l = ain()
        d = deque([])
        p = 1
        for i in range(n):
            if i == 0:
                if l[i] == 0:
                    d.append('a')
                    d.append('b')
                    p = 3
                else:
                    d.append('a'*l[i])
                    d.append('a'*l[i])
                    p = 2
            else:
                if l[i] == 0:
                    d.append(chr(p+96))
                    p = p+1
                    if p == 27:
                        p = 1
                else:
                    x = len(d[-1])
                    if x >= l[i]:
                        d.append(d[-1][:l[i]])
                    else:
                        d[-1] = d[-1] + chr(p+96)*(l[i]-x)
                        d.append(d[-1])
                        p = (p+1)
                        if p == 27:
                            p = 1
        for i in d:
            print(i)















        






# Fast IO Template starts

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        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")

if os.getcwd() == 'D:\\code':
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
    input = lambda: sys.stdin.readline().rstrip("\r\n")
# Fast IO Template ends
if __name__ == "__main__":
    main()