#include <bits/stdc++.h>
using namespace std;
void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long pow(long long n, long long m) {
  if (m % 2 == 0) {
    long long pro = pow(n, m / 2);
    return ((pro) % 1000000007 * (pro) % 1000000007) % 1000000007;
  } else {
    return ((n) % 1000000007 * (pow(n, m - 1)) % 1000000007) % 1000000007;
  }
}
long long findCountDiff(string str) {
  long long c1 = 0, c2 = 0;
  for (long long i = 0; i < str.length() - 1; i++) {
    if (str[i] == 'a' && str[i + 1] == 'b')
      c1++;
    else if (str[i] == 'b' && str[i + 1] == 'a')
      c2++;
  }
  return abs(c1 - c2);
}
int32_t main() {
  boost();
  long long t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    if (str[0] == str[str.length() - 1]) {
      cout << str << '\n';
    } else {
      if (str[0] == 'a')
        str[0] = 'b';
      else
        str[0] = 'a';
      cout << str << '\n';
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long pow(long long n, long long m) {
  if (m % 2 == 0) {
    long long pro = pow(n, m / 2);
    return ((pro) % 1000000007 * (pro) % 1000000007) % 1000000007;
  } else {
    return ((n) % 1000000007 * (pow(n, m - 1)) % 1000000007) % 1000000007;
  }
}
long long findCountDiff(string str) {
  long long c1 = 0, c2 = 0;
  for (long long i = 0; i < str.length() - 1; i++) {
    if (str[i] == 'a' && str[i + 1] == 'b')
      c1++;
    else if (str[i] == 'b' && str[i + 1] == 'a')
      c2++;
  }
  return abs(c1 - c2);
}
int32_t main() {
  boost();
  long long t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    if (str[0] == str[str.length() - 1]) {
      cout << str << '\n';
    } else {
      if (str[0] == 'a')
        str[0] = 'b';
      else
        str[0] = 'a';
      cout << str << '\n';
    }
  }
  return 0;
}

```