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;
  template <typename t1, typename t2>
  using umap = unordered_map<t1, t2>;
  template <typename t>
  using uset = unordered_set<t>;
  struct pair_hash {
    template <class T1, class T2>
    std::size_t operator()(const std::pair<T1, T2>& p) const {
      auto h1 = std::hash<T1>{}(p.first);
      auto h2 = std::hash<T2>{}(p.second);
      return h1 ^ h2;
    }
  };
  template <class Container>
  void split3(const std::string& str, Container& cont, char delim = ' ') {
    std::size_t current, previous = 0;
    current = str.find(delim);
    while (current != std::string::npos) {
      cont.push_back(str.substr(previous, current - previous));
      previous = current + 1;
      current = str.find(delim, previous);
    }
    cont.push_back(str.substr(previous, current - previous));
  }
  long long gcd(long long a, long long b) {
    while (b) {
      a %= b;
      swap(a, b);
    }
    return a;
  }
  void zf(string& s, vector<long long>& z) {
    long long l = 0, r = 0;
    for (long long i = 1; i < s.size(); i++) {
      if (r >= i) {
        z[i] = min(z[i - l], r - i + 1);
      }
      while (z[i] + i < s.size() && s[z[i]] == s[z[i] + i]) {
        z[i]++;
      }
      if (i + z[i] - 1 > r) {
        l = i;
        r = i + z[i] - 1;
      }
    }
  }
  vector<long long> zf(string& a, string& b, string delimiter = "$") {
    string s = b + delimiter + a;
    vector<long long> z(s.size(), 0);
    long long l = 0, r = 0;
    for (long long i = 1; i < s.size(); i++) {
      if (r >= i) {
        z[i] = min(z[i - l], r - i + 1);
      }
      while (z[i] + i < s.size() && s[z[i]] == s[z[i] + i]) {
        z[i]++;
      }
      if (i + z[i] - 1 > r) {
        l = i;
        r = i + z[i] - 1;
      }
    }
    return z;
  }
  const long long INF = 1e9 + 1;
  const long long MOD = 1e9 + 7;
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    long long t;
    cin >> t;
    cin.ignore();
    for (long long _ = 0; _ < t; _++) {
      string s;
      getline(cin, s);
      long long ab = 0, ba = 0;
      for (long long i = 0; i < s.size() - 1; i++) {
        if (s[i] == 'a' && s[i + 1] == 'b') {
          ab++;
        }
        if (s[i] == 'b' && s[i + 1] == 'a') {
          ba++;
        }
      }
      if (ba == ab) {
        cout << s << "\n";
      } else if (ba > ab) {
        cout << 'a' << s.substr(1) << "\n";
      } else {
        cout << 'b' << s.substr(1) << "\n";
      }
    }
    return 0;
  }
  