#include <bits/stdc++.h>
using namespace std;
int solveSimple(string s) {
  int ab = 0, ba = 0;
  int l = s.length();
  for (int i = 0; i < l - 1; i++)
    if (s[i] != s[i + 1])
      if (s[i] == 'a')
        ab++;
      else
        ba++;
  return abs(ab - ba);
}
char flip(char a) {
  if (a == 'a') return 'b';
  return 'a';
}
void solve() {
  string s;
  cin >> s;
  int l = s.length();
  vector<string> helpers;
  helpers.push_back(s);
  string t = s;
  t[0] = flip(t[0]);
  helpers.push_back(t);
  t[0] = flip(t[0]);
  t[l - 1] = flip(t[l - 1]);
  helpers.push_back(t);
  t[0] = flip(t[0]);
  helpers.push_back(t);
  for (unsigned int i = 0; i < helpers.size(); i++) {
    if (solveSimple(helpers[i]) == 0) {
      cout << helpers[i] << endl;
      return;
    }
  }
}
int main() {
  int n;
  cin >> n;
  while (n--) solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
int solveSimple(string s) {
  int ab = 0, ba = 0;
  int l = s.length();
  for (int i = 0; i < l - 1; i++)
    if (s[i] != s[i + 1])
      if (s[i] == 'a')
        ab++;
      else
        ba++;
  return abs(ab - ba);
}
char flip(char a) {
  if (a == 'a') return 'b';
  return 'a';
}
void solve() {
  string s;
  cin >> s;
  int l = s.length();
  vector<string> helpers;
  helpers.push_back(s);
  string t = s;
  t[0] = flip(t[0]);
  helpers.push_back(t);
  t[0] = flip(t[0]);
  t[l - 1] = flip(t[l - 1]);
  helpers.push_back(t);
  t[0] = flip(t[0]);
  helpers.push_back(t);
  for (unsigned int i = 0; i < helpers.size(); i++) {
    if (solveSimple(helpers[i]) == 0) {
      cout << helpers[i] << endl;
      return;
    }
  }
}
int main() {
  int n;
  cin >> n;
  while (n--) solve();
  return 0;
}

```