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: a=eval(input())
  e=[]
  def str_count(a,b):
  	length_a = len(a)
  	length_b = len(b)
  	count = 0
  	i = 0
  	for j in range(length_a):
  		if a[j] == b[i]:
  			if i == length_b-1:
  				count = count+1
  				i = 0
  			else:
  				i += 1
  				if j<(length_a-1) and a[j+1] != b[i]:
  					i = 0
  			
  		else:
  			continue
  	return count
  
  def judge(j):
      cnt1=str_count(j,"ab")
      cnt2=str_count(j,"ba")
      while(cnt1 != cnt2):
          if cnt1 > cnt2:
              b=j.find('ab',0,len(j))
              j=list(j)
              if j[len(j)-1] == 'b':
                  j[len(j)-1] = 'a'
              else:
                  j[b]='b'
              str1="".join(j)
              cnt1=str_count(str1,"ab")
              cnt2=str_count(str1,"ba")
              j=str1
          elif cnt2 > cnt1:
              b=j.find('ba',0,len(j))
              j=list(j)
              if j[len(j)-1] == 'a':
                  j[len(j)-1] = 'b'
              else:
                  j[b]='a'
              str1="".join(j)
              cnt1=str_count(str1,"ab")
              cnt2=str_count(str1,"ba")
              j=str1
      return j
  for i in range(a):
      j=input()
      j=judge(j)
      e.append(j)
  for i in e:
      print(i)
  