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;
  void setIO(string name = "") {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    if (int((name).size())) {
      freopen((name + ".in").c_str(), "r", stdin);
      freopen((name + ".out").c_str(), "w", stdout);
    }
  }
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  long long mcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  bool prime(long long n, long long i) {
    if (i > sqrt(n)) return true;
    if (n % i == 0)
      return false;
    else
      return prime(n, i + 1);
  }
  struct compii {
    bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
      if (a.first == a.second) return a.second < b.second;
      return a.first > b.first;
    }
  };
  bool comp(int a, int b) { return a > b; }
  namespace operators {
  template <typename T1, typename T2>
  istream& operator>>(istream& in, pair<T1, T2>& x) {
    in >> x.first >> x.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream& operator<<(ostream& out, pair<T1, T2> x) {
    out << x.first << " " << x.second;
    return out;
  }
  template <typename T1>
  istream& operator>>(istream& in, vector<T1>& x) {
    for (auto& i : x) in >> i;
    return in;
  }
  template <typename T1>
  ostream& operator<<(ostream& out, vector<T1>& x) {
    for (auto& i : x) out << i << " ";
    return out;
  }
  template <typename T1, typename T2>
  ostream& operator<<(ostream& out, vector<pair<T1, T2>>& x) {
    for (auto& i : x) out << i.first << " " << i.second;
    return out;
  }
  template <typename T1, typename T2>
  istream& operator>>(istream& in, vector<pair<T1, T2>>& x) {
    for (auto& i : x) in >> i.first >> i.second;
    return in;
  }
  }  // namespace operators
  using namespace operators;
  int dx[] = {1, 0, -1, 0};
  int dy[] = {0, 1, 0, -1};
  const int mod = 1e9 + 7;
  const long long inf = 1e18;
  void solve() {
    string second;
    cin >> second;
    int n = int((second).size());
    int c1 = 0, c2 = 0;
    for (int i = 0; i < n - 1; i++) {
      string aux = "";
      aux += second[i];
      aux += second[i + 1];
      if (aux == "ab") c1++;
    }
    for (int i = 0; i < n - 1; i++) {
      string aux = "";
      aux += second[i];
      aux += second[i + 1];
      if (aux == "ba") c2++;
    }
    if (c1 == c2) {
      cout << second << endl;
      return;
    } else {
      second[0] = 'b';
      second[n - 1] = 'b';
      cout << second << endl;
    }
  }
  int main() {
    setIO("");
    int t = 1;
    cin >> t;
    while (t-- > 0) solve();
    return 0;
  }
  