#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.length();
    long long c1 = 0, c2 = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (s[i] == 'a' && s[i + 1] == 'b')
        c1++;
      else if (s[i] == 'b' && s[i + 1] == 'a')
        c2++;
    }
    string ans = "s";
    if (c1 != c2) {
      if (c1 < c2) {
        for (long long i = 0; i < n; i++) {
          if (s[i] == 'b') {
            s[i] = 'a';
            break;
          }
        }
        cout << s << endl;
      } else if (c1 > c2) {
        for (long long i = 0; i < n; i++) {
          if (s[i] == 'a') {
            s[i] = 'b';
            break;
          }
        }
        cout << s << "\n";
      }
    } else
      cout << s << endl;
  }
  return 0;
}

### 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;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.length();
    long long c1 = 0, c2 = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (s[i] == 'a' && s[i + 1] == 'b')
        c1++;
      else if (s[i] == 'b' && s[i + 1] == 'a')
        c2++;
    }
    string ans = "s";
    if (c1 != c2) {
      if (c1 < c2) {
        for (long long i = 0; i < n; i++) {
          if (s[i] == 'b') {
            s[i] = 'a';
            break;
          }
        }
        cout << s << endl;
      } else if (c1 > c2) {
        for (long long i = 0; i < n; i++) {
          if (s[i] == 'a') {
            s[i] = 'b';
            break;
          }
        }
        cout << s << "\n";
      }
    } else
      cout << s << endl;
  }
  return 0;
}

```