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: def main():
      alpha = 'abcdefghijklmnopqrstuvwxyz'
      ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
      inf = 1e17
      # mod = 998244353
      mod = 10 ** 9 + 7
  
      # Max = 10 ** 6
      # primes = []
      # prime = [True for i in range(Max + 1)]
      # p = 2
      # while (p * p <= Max + 1):
      #
      #     # If prime[p] is not
      #     # changed, then it is a prime
      #     if (prime[p] == True):
      #
      #         # Update all multiples of p
      #         for i in range(p * p, Max + 1, p):
      #             prime[i] = False
      #     p += 1
      #
      # for p in range(2, Max + 1):
      #     if prime[p]:
      #         primes.append(p)
  
      # print(len(primes))
  
      def factorial(n):
          f = 1
  
          for i in range(1, n + 1):
              f = (f * i) % mod  # Now f never can
          # exceed 10^9+7
  
          return f
  
      def factorial_by(n, by):
          f = 1
  
          for i in range(1, n + 1):
              if i == by:
                  continue
              f = (f * i) % mod  # Now f never can
          # exceed 10^9+7
  
          return f
  
      def ncr(n, r):
          # initialize numerator
          # and denominator
          num = den = 1
          for i in range(r):
              num = (num * (n - i)) % mod
              den = (den * (i + 1)) % mod
          return (num * pow(den,
                            mod - 2, mod)) % mod
  
      def primeFactors(num):
  
          pf = []
          while num % 2 == 0:
              pf.append(2)
              num = num // 2
  
          # n must be odd at this point
          # so a skip of 2 ( i = i + 2) can be used
          for i in range(3, int(math.sqrt(num)) + 1, 2):
  
              # while i divides n , print i and divide n
              while num % i == 0:
                  pf.append(i)
                  num = num // i
  
          # Condition if n is a prime
          # number greater than 2
          if num > 2:
              pf.append(num)
  
          return pf
  
      class Node(object):
          def __init__(self, name, d):
              self.d = d
              self.name = name
  
          def __repr__(self):
              pass
  
      def solve(s):
          s_str = "".join(s)
          if s_str.count('ab') > s_str.count('ba'):
              if s[0] == 'a':
                  s[0] = 'b'
  
              else:
                  s[-1] = 'b'
          elif s_str.count('ab') < s_str.count('ba'):
  
              if s[0] == 'b':
                  s[0] = 'a'
              else:
                  s[-1] = 'a'
  
  
          return "".join(s)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
      t = int(input())
      ans = []
      for _ in range(t):
          s = list(input()[:-1])
          #n = int(input())
  
          # n,m,k= map(int, input().split())
          # grid = []
          # for i in range(n):
          #     grid.append([int(x) for x in input().split()])
          # #s = list(map(int,list(input()[:-1])))
          #a = [int(x) for x in input().split()]
          # b = [int(x) for x in input().split()]
          # q = int(input())
          # queries = []
          # for i in range(q):
          #     x1,x2 = map(int, input().split())
          #     queries.append([x1,x2])
  
          # tree = []
          #
          #
          # for i in range(n+1):
          #     tree.append([])
          #
          # for i in range(n-1):
          #     u, v = map(int, input().split())
          #     tree[u].append(v)
          #     tree[v].append(u)
          #
          ans.append(solve(s))
  
      p = 1
      for answer in ans:
          # print('Case #'+str(p)+": "+str(answer))
          print(answer)
          p += 1
  
  
  if __name__ == "__main__":
      import sys, threading
      import bisect
      import math
      import itertools
      from sys import stdout
  
      ############  Sorted Containers  ######################
      import heapq
      from queue import PriorityQueue
      from collections import deque
  
      ############ Tree Problems ( Use Python 3) ###########
      #sys.setrecursionlimit(2 ** 32 // 2 - 1)
      #threading.stack_size(1 << 27)
  
      input = sys.stdin.readline
      thread = threading.Thread(target=main)
      thread.start()
      thread.join()
  