#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    auto count = [&n](const string &s) {
      int d = 0;
      for (int i = 0; i < n - 1; i++) {
        if (s[i] == 'a' && s[i + 1] == 'b')
          d++;
        else if (s[i] == 'b' && s[i + 1] == 'a')
          d--;
      }
      return d;
    };
    while (1) {
      int d = count(s);
      if (count(s) == 0) break;
      int vv = abs(d);
      for (int i = 0; i < n - 1; i++) {
        char tmp = s[i];
        if (s[i] == 'a')
          s[i] = 'b';
        else
          s[i] = 'a';
        int vvv = abs(count(s));
        if (vvv < vv) {
          break;
        }
        s[i] = tmp;
      }
    }
    cout << s << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
using int64 = long long;
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    auto count = [&n](const string &s) {
      int d = 0;
      for (int i = 0; i < n - 1; i++) {
        if (s[i] == 'a' && s[i + 1] == 'b')
          d++;
        else if (s[i] == 'b' && s[i + 1] == 'a')
          d--;
      }
      return d;
    };
    while (1) {
      int d = count(s);
      if (count(s) == 0) break;
      int vv = abs(d);
      for (int i = 0; i < n - 1; i++) {
        char tmp = s[i];
        if (s[i] == 'a')
          s[i] = 'b';
        else
          s[i] = 'a';
        int vvv = abs(count(s));
        if (vvv < vv) {
          break;
        }
        s[i] = tmp;
      }
    }
    cout << s << "\n";
  }
  return 0;
}

```