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: num_testcases = int(input(""));
  lis = [0] * num_testcases;
  
  for j in range(num_testcases):
  	lis[j] = input("");
  
  def ab_balance(stra):
  	num_ab = 0;
  	num_ba = 0;
  
  	pos_ab = 0;
  	pos_ba = 0;
  
  	boolean1 = True;
  	boolean2 = True;
  	n1 = 0;
  	n2 = 0;
  	while boolean1:
  		if stra.find("ab",pos_ab) != -1 and pos_ab < len(stra) - 1: 
  			num_ab = num_ab + 1;
  			pos_ab = stra.find("ab",pos_ab) + 2;
  		else:
  			boolean1 = False;
  	while boolean2:
  		if stra.find("ba",pos_ba) != -1 and pos_ba < len(stra) - 1: 
  			# print("pos_ba is ",pos_ba);
  
  			num_ba = num_ba + 1;
  			pos_ba = stra.find("ba",pos_ba) + 2;
  		else:
  			boolean2 = False;
  	# print(num_ab, "    ba is: ", num_ba);
  	if num_ab == num_ba:
  		print(stra);
  	elif num_ab < num_ba:
  		print(stra[:len(stra)-1] + "b");
  	elif num_ba < num_ab:
  		print(stra[:len(stra)-1] + "a");
  for i in range(num_testcases):
  	ab_balance(lis[i]);