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 bisect
  import copy
  import decimal
  import fractions
  import heapq
  import itertools
  import math
  import random
  import sys
  from collections import Counter,deque,defaultdict
  from functools import lru_cache,reduce
  from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
  def _heappush_max(heap,item):
      heap.append(item)
      heapq._siftdown_max(heap, 0, len(heap)-1)
  def _heappushpop_max(heap, item):
      if heap and item < heap[0]:
          item, heap[0] = heap[0], item
          heapq._siftup_max(heap, 0)
      return item
  from math import gcd as GCD, modf
  read=sys.stdin.read
  readline=sys.stdin.readline
  readlines=sys.stdin.readlines
  
  t=int(readline())
  for _ in range(t):
      N=int(readline())
      if N==1:
          ans='a'
      elif N%2==0:
          ans='a'*(N//2)+'b'+'a'*(N//2-1)
      else:
          ans='a'*(N//2)+'bc'+'a'*(N//2-1)
      print(ans)