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>
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    std::cerr << name << " : " << arg1 << "\n";
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                            std::pair<long long, long long> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
    in >> a.first >> a.second;
    return in;
  }
  std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
    in >> a.first >> a.second;
    return in;
  }
  long long bpow(long long x, long long v) {
    if (v == 0) return 1 % 1000000007;
    if (v % 2 == 1) return (x * bpow(x, v - 1)) % 1000000007;
    long long r = bpow(x, v / 2);
    return (r * r) % 1000000007;
  }
  using namespace std;
  void solve() {
    string str, s;
    cin >> str;
    vector<int> v;
    for (int i = 0; i < str.size();) {
      char ch = str[i];
      int c = 0;
      while (i < str.size() && str[i] == ch) {
        c++;
        i++;
      }
      s += ch;
      v.push_back(c);
    }
    if (s[0] == s[s.size() - 1])
      cout << str << "\n";
    else {
      if (str[0] == 'a')
        str[0] = 'b';
      else
        str[0] = 'a';
      cout << str << "\n";
    }
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  