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 collections import deque
  
  def get_count(string, x, y):
      count = 0
      for i in range(len(string)-1):
          if string[i] == x and string[i+1] == y:
              count += 1
      return count
          
  
  def solve():
      string = input()
      ab, ba = get_count(string, "a", "b"), get_count(string, "b", "a")
      if ab == ba:
          return string
      
      q, visited = deque(), set()
      q.append(string)
      
      while q:
          # print("q", q)
          for _ in range(len(q)):
              # print("q", q)
              cur = q.popleft()
              # print("cur->", cur, "visited->", visited)
              # print("cur", cur)
              if cur in visited:
                  # print("cur is visited")
                  continue
              else:
                  # print("cur is NOT visited")
                  visited.add(cur)
                  # print("visited now", visited)
              
              ab, ba = get_count(cur, "a", "b"), get_count(cur, "b", "a")
              # print("ab, ba", ab, ba)
              if ab == ba:
                  return cur
                
              for i in range(len(cur)):
                  char = string[i]
                  if char == "a":
                      new_char = "b"
                  else:
                      new_char = "a"
                  
                  new_string = string[:i] + new_char + string[i+1:]
                  # print("new_string", new_string)
                  if new_string in visited:
                      # print("new_string is visited", new_string)
                      continue
                  # else:
                  q.append(new_string)
          # print("\n")
      
      
  for i in range(int(input())):
      print(solve())