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: #include <bits/stdc++.h>
  using namespace std;
  long long gcd(long long a, long long b) {
    if (a % b == 0) {
      return b;
    } else {
      return gcd(b, a % b);
    }
  }
  long long powc(long long a, long long b) {
    if (b < 0) {
      return 0;
    }
    long long md = b;
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res *= a;
      }
      a *= a;
      b >>= 1;
    }
    return res;
  }
  bool isPowerOfTwo(long long n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  bool valid(long long i, long long j, long long n, long long m) {
    if (i >= 0 && i < n && j >= 0 && j < m) return true;
    return false;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      long long x = 0, y = 0;
      for (long long i = 1; i < s.size(); i++) {
        if (s[i] == 'b' && s[i - 1] == 'a') {
          x++;
        }
        if (s[i] == 'a' && s[i - 1] == 'b') {
          y++;
        }
      }
      if (x == y) {
        cout << s << "\n";
      } else {
        if (s[0] == 'a') {
          s[s.size() - 1] = 'a';
        } else {
          s[s.size() - 1] = 'b';
        }
        cout << s << "\n";
      }
    }
  }
  