#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &x : v) in >> x;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &x : v) out << x << ' ';
  return out;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
const bool tests = 1;
void solve_case() {
  string s;
  cin >> s;
  long long n = s.length();
  if (s[0] != s[n - 1]) {
    s[n - 1] = s[0];
  }
  cout << s << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long testcase = 1;
  if (tests) cin >> testcase;
  cout << fixed << setprecision(10);
  for (long long test = 1; test <= testcase; test++) {
    solve_case();
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &x : v) in >> x;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &x : v) out << x << ' ';
  return out;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
const bool tests = 1;
void solve_case() {
  string s;
  cin >> s;
  long long n = s.length();
  if (s[0] != s[n - 1]) {
    s[n - 1] = s[0];
  }
  cout << s << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long testcase = 1;
  if (tests) cin >> testcase;
  cout << fixed << setprecision(10);
  for (long long test = 1; test <= testcase; test++) {
    solve_case();
  }
  return 0;
}

```