description: You are given an integer n. Find any string s of length n consisting only of English lowercase letters such that each non-empty substring of s occurs in s an odd number of times. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.
  
  A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print a single line containing the string s. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.
  
  Example
  
  Input
  
  
  4
  3
  5
  9
  19
  
  
  Output
  
  
  abc
  diane
  bbcaabbba
  youarethecutestuwuu
  
  Note
  
  In the first test case, each substring of "abc" occurs exactly once.
  
  In the third test case, each substring of "bbcaabbba" occurs an odd number of times. In particular, "b" occurs 5 times, "a" and "bb" occur 3 times each, and each of the remaining substrings occurs exactly once.
solution: # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// PYTHON IS THE BEST ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  import sys,os,io
  from sys import stdin
  import math 
  from collections import defaultdict
  from heapq import heappush, heappop, heapify
  from bisect import bisect_left , bisect_right
  from io import BytesIO, IOBase
  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")
  
  
  
  
  alphabets = list('abcdefghijklmnopqrstuvwxyz')
  
  
  
  #for deep recursion__________________________________________-
  from types import GeneratorType
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to = stack[-1].send(to)
              return to
  
      return wrappedfunc
  
  def ncr(n, r, p):  
      num = den = 1
      for i in range(r):
          num = (num * (n - i)) % p
          den = (den * (i + 1)) % p
      return (num * pow(den,p - 2, p)) % p
  
  def primeFactors(n): 
      l = []
      while n % 2 == 0: 
          l.append(2)
          n = n / 2
      for i in range(3,int(math.sqrt(n))+1,2): 
          while n % i== 0: 
              l.append(int(i))
              n = n / i 
      if n > 2: 
          l.append(n)
      # c = dict(Counter(l))
      return list(set(l))
      # return c
  
  def power(x, y, p) : 
  	res = 1
  	x = x % p 
  	if (x == 0) : 
  		return 0
  	while (y > 0) : 
  		if ((y & 1) == 1) : 
  			res = (res * x) % p 
  		y = y >> 1	 # y = y/2 
  		x = (x * x) % p 		
  	return res 
  
  #____________________GetPrimeFactors in log(n)________________________________________
  def sieveForSmallestPrimeFactor():
      MAXN = 100001
      spf = [0 for i in range(MAXN)]
      spf[1] = 1
      for i in range(2, MAXN):
          spf[i] = i
      for i in range(4, MAXN, 2):
          spf[i] = 2
      for i in range(3, math.ceil(math.sqrt(MAXN))):
          if (spf[i] == i):
              for j in range(i * i, MAXN, i): 
                  if (spf[j] == j):
                      spf[j] = i
      return spf
  def getPrimeFactorizationLOGN(x):
      spf = sieveForSmallestPrimeFactor()
      ret = list()
      while (x != 1):
          ret.append(spf[x])
          x = x // spf[x]  
      return ret
  #____________________________________________________________
  
  
  
  def SieveOfEratosthenes(n): 
      #time complexity = nlog(log(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
      return prime
  def si():
      return input()
  def divideCeil(n,x):
      if (n%x==0):
          return n//x
      return n//x+1
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  
  
  # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  if(os.path.exists('input.txt')):
      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")
  
  
  def solve():
      n = ii()
      if n<=5:
          print(''.join(alphabets[:n]))
          return 
      if n%2==1:
          x = n//2 
          ans = ['a']*x + ['b','c'] + ['a']*(x-1)
      else:
          x = n//2 
          ans = ['a']*x + ['b'] + ['a']*(x-1)
      print(''.join(ans))
  
  
  
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  