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.
1
abbabababbabababbababbbabbbbabbbabaabaabba
 4
b
aabbbabaa
abbb
abbaab
 1
aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

abbabababbabababbababbbabbbbabbbabaabaabba
 b
aabbbabaa
bbbb
bbbaab
 baaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

#include <bits/stdc++.h>
constexpr bool test_cases = true;
void solve() {
  std::string s;
  std::cin >> s;
  int n = s.length();
  int ab = 0, ba = 0;
  for (int i = 0; i < n; ++i) {
    if (i + 1 < n) {
      if (s[i] == 'a' and s[i + 1] == 'b')
        ab += 1;
      else if (s[i] == 'b' and s[i + 1] == 'a')
        ba += 1;
    }
  }
  if (ab == ba)
    std::cout << s << '\n';
  else {
    if (std::abs(ab - ba) != 1) throw std::runtime_error("retard");
    if (ab > ba) {
      int last = -1;
      for (int i = 0; i < n; ++i)
        if (s[i] == 'b') last = i;
      s[last] = 'a';
    } else {
      int last = -1;
      for (int i = 0; i < n; ++i)
        if (s[i] == 'a') last = i;
      s[last] = 'b';
    }
    std::cout << s << '\n';
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.precision(10);
  std::cerr.precision(10);
  std::cout << std::fixed << std::boolalpha;
  std::cerr << std::fixed << std::boolalpha;
  int32_t cases = 1;
  if (test_cases) std::cin >> cases;
  while (cases--) solve();
  return 0;
}
