description: You are given a string s of length n consisting of characters a and/or b.
  
  Let \operatorname{AB}(s) be the number of occurrences of string ab in s as a substring. Analogically, \operatorname{BA}(s) is the number of occurrences of ba in s as a substring.
  
  In one step, you can choose any index i and replace s_i with character a or b.
  
  What is the minimum number of steps you need to make to achieve \operatorname{AB}(s) = \operatorname{BA}(s)?
  
  Reminder:
  
  The number of occurrences of string d in s as substring is the number of indices i (1 ≤ i ≤ |s| - |d| + 1) such that substring s_i s_{i + 1} ... s_{i + |d| - 1} is equal to d. For example, \operatorname{AB}(aabbbabaa) = 2 since there are two indices i: i = 2 where aabbbabaa and i = 6 where aabbbabaa.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
  
  The first and only line of each test case contains a single string s (1 ≤ |s| ≤ 100, where |s| is the length of the string s), consisting only of characters a and/or b.
  
  Output
  
  For each test case, print the resulting string s with \operatorname{AB}(s) = \operatorname{BA}(s) you'll get making the minimum number of steps.
  
  If there are multiple answers, print any of them.
  
  Example
  
  Input
  
  
  4
  b
  aabbbabaa
  abbb
  abbaab
  
  
  Output
  
  
  b
  aabbbabaa
  bbbb
  abbaaa
  
  Note
  
  In the first test case, both \operatorname{AB}(s) = 0 and \operatorname{BA}(s) = 0 (there are no occurrences of ab (ba) in b), so can leave s untouched.
  
  In the second test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 2, so you can leave s untouched. 
  
  In the third test case, \operatorname{AB}(s) = 1 and \operatorname{BA}(s) = 0. For example, we can change s_1 to b and make both values zero.
  
  In the fourth test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 1. For example, we can change s_6 to a and make both values equal to 1.
solution: from __future__ import division, print_function
  import math
  import sys
  import os
  from io import BytesIO, IOBase
  #from collections import deque, Counter, OrderedDict, defaultdict
  #import heapq
  #ceil,floor,log,sqrt,factorial,pow,pi,gcd
  #import bisect
  #from bisect import bisect_left,bisect_right
  
  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")
  
  
  def print(*args, **kwargs):
  	"""Prints the values to a stream, or to sys.stdout by default."""
  	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
  	at_start = True
  	for x in args:
  		if not at_start:
  			file.write(sep)
  		file.write(str(x))
  		at_start = False
  	file.write(kwargs.pop("end", "\n"))
  	if kwargs.pop("flush", False):
  		file.flush()
  
  
  if sys.version_info[0] < 3:
  	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
  	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  def inp():
      return(int(input()))
  def inps():
      return input().strip()
  def inlt():
      return(list(map(int,input().split())))
  def insr():
      s = input().strip()
      return(list(s[:len(s)]))
  def invr():
      return(map(int,input().split()))
  def cntab(s):
      cnt=0
      for i in range(len(s)-1):
          if s[i]=='a' and s[i+1]=='b':
              cnt+=1
      return cnt
  
  def cntba(s):
      cnt=0
      for i in range(len(s)-1):
          if s[i]=='b' and s[i+1]=='a':
              cnt+=1
      return cnt
  
  for _ in range(inp()):
      s=insr()
      if cntab(s)==cntba(s):
          print("".join(s))
      else:
          cng={'a':'b','b':'a'}
          for i in range(len(s)):
              g=s.copy()
              g[i]=cng[g[i]]
              v1=cntab(g)
              v2=cntab(g)
              if v1==v2:
                  s=g
                  break
          print("".join(s))