#include <bits/stdc++.h>
using namespace std;
void solve() {
  string s;
  cin >> s;
  long long ct = 0;
  for (int i = 0; i < s.size() - 1; i++) {
    if (s[i] == 'a' && s[i + 1] == 'b')
      ct++;
    else if (s[i] == 'b' && s[i + 1] == 'a')
      ct--;
  }
  if (ct == 0) {
    cout << s << endl;
  } else if (ct == 1) {
    for (int i = 0; i < s.size() - 1; i++) {
      cout << s[i];
    }
    cout << "a\n";
  } else {
    for (int i = 0; i < s.size() - 1; i++) {
      cout << s[i];
    }
    cout << "b\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in CPP 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;
void solve() {
  string s;
  cin >> s;
  long long ct = 0;
  for (int i = 0; i < s.size() - 1; i++) {
    if (s[i] == 'a' && s[i + 1] == 'b')
      ct++;
    else if (s[i] == 'b' && s[i + 1] == 'a')
      ct--;
  }
  if (ct == 0) {
    cout << s << endl;
  } else if (ct == 1) {
    for (int i = 0; i < s.size() - 1; i++) {
      cout << s[i];
    }
    cout << "a\n";
  } else {
    for (int i = 0; i < s.size() - 1; i++) {
      cout << s[i];
    }
    cout << "b\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```