#include <bits/stdc++.h>
using namespace std;
const int nmax = 1e5 + 5;
int t, n;
string s;
bool ktra(string s) {
  int d = 0;
  for (int i = 1; i < n; i++) {
    if (s[i] == 'a' && s[i + 1] == 'b') d++;
    if (s[i] == 'b' && s[i + 1] == 'a') d--;
  }
  return (d == 0);
}
void solve() {
  if (ktra(s)) {
    for (int i = 1; i <= n; i++) cout << s[i];
    cout << '\n';
  } else {
    for (int i = 1; i <= n; i++) {
      string res = s;
      res[i] = (s[i] == 'a' ? 'b' : 'a');
      if (ktra(res)) {
        for (int j = 1; j <= n; j++) cout << res[j];
        cout << '\n';
        return;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> s;
    n = s.size();
    s = ' ' + s;
    solve();
  }
}

### 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;
const int nmax = 1e5 + 5;
int t, n;
string s;
bool ktra(string s) {
  int d = 0;
  for (int i = 1; i < n; i++) {
    if (s[i] == 'a' && s[i + 1] == 'b') d++;
    if (s[i] == 'b' && s[i + 1] == 'a') d--;
  }
  return (d == 0);
}
void solve() {
  if (ktra(s)) {
    for (int i = 1; i <= n; i++) cout << s[i];
    cout << '\n';
  } else {
    for (int i = 1; i <= n; i++) {
      string res = s;
      res[i] = (s[i] == 'a' ? 'b' : 'a');
      if (ktra(res)) {
        for (int j = 1; j <= n; j++) cout << res[j];
        cout << '\n';
        return;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> s;
    n = s.size();
    s = ' ' + s;
    solve();
  }
}

```