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;
  int MOD = 1e9 + 7;
  long long int power(long long int a, long long int b) {
    long long int res = 1;
    a = a % MOD;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % MOD;
        b--;
      }
      a = (a * a) % MOD;
      b >>= 1;
    }
    return res;
  }
  long long int fermat_inv(long long int y) { return power(y, MOD - 2); }
  long long int gcd(long long int a, long long int b) {
    return (b == 0) ? a : gcd(b, a % b);
  }
  long long int min(long long int a, long long int b) { return (a > b) ? b : a; }
  long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
  bool prime[1000001];
  vector<int> primes;
  void SieveOfEratosthenes(int n) {
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = 0;
    for (int p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (int i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    for (int p = 2; p < 1000001; p++)
      if (prime[p]) primes.push_back(p);
  }
  long long int fact[1000010];
  long long int finv[1000010];
  void factorial(int n) {
    fact[0] = 1;
    finv[0] = 1;
    for (int i = 1; i <= n; i++)
      fact[i] = fact[i - 1] * i, fact[i] %= MOD, finv[i] = fermat_inv(fact[i]);
  }
  long long int ncr(long long int n, long long int r) {
    if (n < r)
      return 0;
    else {
      long long int x = finv[r] * finv[n - r] % MOD;
      return fact[n] * x % MOD;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int te = 1;
    cin >> te;
    while (te--) {
      string s;
      cin >> s;
      int n = s.size();
      s[n - 1] = s[0];
      cout << s << '\n';
    }
  }
  