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: from math import *
  from decimal import *
  from decimal import Decimal as D
  from string import ascii_lowercase as lc
  from string import ascii_uppercase as uc
  from itertools import *
  from itertools import permutations as perm
  from itertools import product as prod
  from sys import *
  ip = input
  ra = lambda: list(map(int, input().split()))
  ri = lambda: int(input())
  rd = lambda: map(int, input().split())
  MOD = 10 ** 9 + 7
  MOD2 = 998244353
  def debug(*a):
      stderr.write(' '.join(map(str, a)) + '\n')
  def fp(x, p, m):
      r = 1
      while p:
          if p % 2:
              r = r * x % m
              p -= 1
          else:
              x = x * x % m
              p //= 2
      return r
  def fact(x, m):
      r = 1
      for i in range(2, x + 1):
          r = r * i % m
      return r
  def C(n, k, m):
      return fact(n, m) * fp(fact(n - k, m), m - 2, m) % m * fp(fact(k, m), m - 2, m) % m
  class SegmentTree:
      def f(self, a, b):
          return a + b
      def __init__(self, n, a):
          self.tree = [0] * (2 * n)
          self.n = n
          for i in range(n):
              self.tree[n + i] = a[i]
          for i in range(n - 1, 0, -1):
              self.tree[i] = self.f(self.tree[2 * i], self.tree[2 * i + 1])
      def q(self, l, r):
          l += self.n
          r += self.n
          res = 0
          while r > 1:
              if l % 2:
                  res = self.f(res, self.tree[l]) 
                  l += 1
              if r % 2:
                  r -= 1
                  res = self.f(res, self.tree[r])
              r //= 2
              l //= 2
          return res
      def u(self, index, value):
          index += self.n
          self.tree[index] = value
          index //= 2
          while index > 0:
              self.tree[index] = self.f(self.tree[2 * index], self.tree[2 * index + 1])
              index //= 2
  def z_function(s):
      l, r, n = 0, 0, len(s)
      z = [0] * n
      z[0] = 0
      for i in range(1, n):
          if r >= i: z[i] = min(z[i - l], r - i + 1)
          while z[i] + i < n and s[z[i]] == s[z[i] + i]: z[i] += 1
          if i + z[i] - 1 > r:
              l = i
              r = l + z[i] - 1
      return z
  def pc_pre(a):
      n = len(a)
      r = [0] * (n + 1)
      for i in range(n):
          r[i + 1] = r[i] + a[i]
      return r
  def pc_suf(a):
      n = len(a)
      r = [0] * (n + 1)
      for i in range(n - 1, -1, -1):
          r[i] = r[i + 1] + a[i]
      return r
  def bsloe(a, v):
      l, h = -1, len(a)
      while h - l > 1:
          m = (h + l) // 2
          if a[m] > v: h = m
          else: l = m
      return l
  def bsh(a, v):
      l, h = -1, len(a)
      while h - l > 1:
          m = (h + l) // 2
          if a[m] > v: h = m
          else: l = m
      return h
  def test():
      a = [1, 2, 3]
      s = 'abcabcab'
      assert(C(3, 2, 5) == 3)
      assert(pc_pre(a) == [0, 1, 3, 6])
      assert(pc_suf(a) == [6, 5, 3, 0])
      assert(z_function(s) == [0, 0, 0, 5, 0, 0, 2, 0])
      assert(fp(2, 3, 5) == 3)
      assert(fact(3, 5) == 1)
      t = SegmentTree(3, a)
      assert(t.q(1, 3) == t.f(a[1], a[2]))
      t.u(0, 5)
      assert(t.q(0, 2) == t.f(5, a[1]))
      assert(bsh(a, 3) == 3)
      assert(bsloe(a, 3) == 2)
      debug("tests passed")
  test()
  ################    code goes here    ################
  def solve(n):
      if n == 1:
          print('a')
      elif n % 2 == 0:
          print('a' * (n // 2) + 'b' + 'a' * (n//2-1))
      else:
          print('a' * (n // 2) + 'bc' + 'a' * (n//2-1))
      
  t, = rd()
  for _ in range(t):
      n, = rd()
      solve(n)