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>
  int dx[8] = {0, 1};
  int dy[8] = {1, 0};
  long long gcd(long long x, long long y) { return (!y ? x : gcd(y, x % y)); }
  long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
  using namespace std;
  const long long LLOO = 0x3f3f3f3f3f3f3f3f;
  const long long OO = 0x3f3f3f3f;
  const long double ESP = 1e-8;
  void fast() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int n;
  vector<long long> arr, bit;
  void add(int indx, int toadd) {
    for (int i = indx; i <= n; i += (i & -i)) {
      if (arr[i] > toadd) bit[i] += toadd;
    }
  }
  long long getprfix(long long indx) {
    int sum = 0;
    for (int i = indx; i > 0; i -= (i & -i)) {
      sum += bit[i];
    }
    return sum;
  }
  long long rangesum(int l, int r) { return (getprfix(r) - getprfix(l - 1)); }
  void buil() {
    bit = vector<long long>(n + 1, 0);
    for (int i = 1; i <= n; i++) add(i, arr[i]);
  }
  int main() {
    fast();
    int t;
    cin >> t;
    while (t--) {
      int ba = 0, ab = 0, c = 0;
      string s;
      cin >> s;
      for (int 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 (ab == ba)
        cout << s << "\n";
      else {
        s[s.size() - 1] = s[0];
        cout << s << "\n";
      }
    }
  }
  