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>
  long long mod = 1000000007;
  using namespace std;
  const int MAX_N = 1e6 + 2, BIT = 34;
  long long poww(long long x, long long n) {
    if (n == 0) return 1;
    if (n == 1) return x % mod;
    long long ans = poww(x, n / 2);
    ans = (ans * ans) % mod;
    if (n % 2) return (ans * x) % mod;
    return ans % mod;
  }
  struct pt {
    long long x, y;
    pt() {}
    pt(long long _x, long long _y) : x(_x), y(_y) {}
    pt operator+(const pt &p) const { return pt(x + p.x, y + p.y); }
    pt operator-(const pt &p) const { return pt(x - p.x, y - p.y); }
    long long cross(const pt &p) const { return x * p.y - y * p.x; }
    long long dot(const pt &p) const { return x * p.x + y * p.y; }
    long long cross(const pt &a, const pt &b) const {
      return (a - *this).cross(b - *this);
    }
    long long dot(const pt &a, const pt &b) const {
      return (a - *this).dot(b - *this);
    }
    long long sqrLen() const { return this->dot(*this); }
    void Out() { cout << x << " " << y << endl; }
  };
  long long Gcd(long long a, long long b) {
    if (b > a) swap(a, b);
    if (b == 0) return a;
    return Gcd(b, a % b);
  }
  long long Pref[MAX_N];
  long long Inv[MAX_N];
  long long A[MAX_N];
  int main() {
    long long Tc = 1;
    cin >> Tc;
    while (Tc--) {
      string s;
      cin >> s;
      long long n = s.size();
      char f = s[0];
      long long fi = 0;
      char e = s[n - 1];
      long long en = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == f)
          fi++;
        else
          break;
      }
      for (int i = n - 1; i >= 0; i--) {
        if (s[i] == e)
          en++;
        else
          break;
      }
      if (f != e) {
        s[0] = e;
      }
      cout << s << endl;
    }
    return 0;
  }
  