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: import os
  import sys
  from io import BytesIO, IOBase
  #import sys
  #input = sys.stdin.buffer.readline
  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")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  #######################################
  from itertools import permutations
  from collections import *
  from collections import deque
  from operator import itemgetter , attrgetter
  from decimal import *
  import bisect
  import math
  import heapq as hq
  #import sympy
  MOD=10**9 +7
  
  
  def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    # since all primes > 3 are of the form 6n ± 1
    # start with f=5 (which is prime)
    # and test f, f+2 for being prime
    # then loop by 6.
    f = 5
    while f <= r:
  
      if n % f == 0: return False
      if n % (f+2) == 0: return False
      f += 6
    return True
  
  def pow(a,b,m):
      ans=1
      while b:
          if b&1:
              ans=(ans*a)%m
          b//=2
          a=(a*a)%m
      return ans
  #vis=[]
  #graph=[]
  
  def ispalindrome(s):
      if s[:]==s[::-1]:
          return 1
      return 0
  dp=[]
  limit=[]
  v=[]
  def dpdfs(u,t=-1):
      dp[0][u]=0
      dp[1][u]=0
      for i in v[u]:
          if i==t:
  
              continue
          if dp[1][i]==-1:
              dpdfs(i,u)
          dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
          dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
  vis=[]
  f=0
  def dfs(i):
      vis[i]=1
      act[i]=1
      for j in v[i]:
          if act[j]:
              f=1
              #print(-1)
              return -1
          if vis[j]==0:
             if dfs(j)==-1:
                 return -1
      act[i]=0
      ans.append(i)
      return 0
  from queue import PriorityQueue
  
  def z_algorithm(s):
      res = [0] * len(s)
      res[0] = len(s)
      i, j = 1, 0
      while i < len(s):
          while i + j < len(s) and s[j] == s[i + j]:
              j += 1
          res[i] = j
          if j == 0:
              i += 1
              continue
          k = 1
          while i + k < len(s) and k + res[k] < j:
              res[i + k] = res[k]
              k += 1
          i, j = i + k, j - k
      return res
  def gcd(a, b):
      if a == 0:
          return b
      return gcd(b % a, a)
  
  
  # Function to return LCM of two numbers
  def lcm(a, b):
      return (a / gcd(a, b)) * b
  
  
  def gcdExtended(a, b):
      # Base Case
      if a == 0:
          return b, 0, 1
  
      gcd, x1, y1 = gcdExtended(b % a, a)
  
      # Update x and y using results of recursive
      # call
      x = y1 - (b // a) * x1
      y = x1
  
      return gcd, x, y
  import re
  for _ in range(int(input())):
      n=int(input())
      ans=""
      if n==1:
          print("a")
          continue
      if n%2:
          ans+="a"*(n//2)
          ans+="bc"
          ans+="a"*(n//2-1)
      else:
          ans+="a"*(n//2)
          ans+="b"
          ans+="a"*(n//2-1)
      print(ans)