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 _typeshed import SupportsKeysAndGetItem
  import sys
  from collections import deque
  import heapq
  import bisect
  import copy
  # sys.stdin=open("input.txt","r");
  # sys.stdout=open("output.txt","w")
  ####### GLOBAL ###############
  MOD=1000000007
  NO=lambda:print("NO")
  YES=lambda:print("YES")
  _1=lambda:print(-1)
  ari=lambda:[int(_) for _ in input().split()]
  cin=lambda:int(input())
  cis=lambda:input()
  show=lambda x: print(x)
  fast= lambda:sys.stdin.readline()
  
  ########### END #########
  ######
  test_case=1
  test_case=int(input())
  ######
  sys.setrecursionlimit(10**6)
      
  def ans():
      s=cis()
      abcount=0
      bacount=0
      n=len(s)
      s=list(s)
      def bruteforce(news):
          abcount=0
          nonlocal n
          bacount=0
          for i in range(n-1):
              if s[i]=="a" and s[i+1]=="b":
                  abcount+=1
              if s[i]=="b" and s[i+1]=="a":
                  bacount+=1
          # print(abcount,bacount)
          if abcount==bacount:
              return "".join(news)
          return False
      ans=bruteforce(s)
      if bruteforce(s)!=False:
          print(ans)
          return 
          
      for i in range(n):
          val=s[i]
          if s[i]=="a":
              s[i]="b"
              ans=bruteforce(s)
              if ans!=False:
                  print(ans)
                  return
              s[i]="a"
          else:
              s[i]="a"
              ans=bruteforce(s)
              if ans!=False:
                  print(ans)
                  return
              s[i]="b"
      print(ans)
              
              
      # print("b"*len(a))
      return
  for _ in range(test_case):
      ans()
      