#include <bits/stdc++.h>
using namespace std;
inline int read() {
  bool ok = 0;
  int res = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    ok = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  return ok ? ~res + 1 : res;
}
int t, n, m, k, x, y, len;
char s[110];
int main() {
  t = read();
  while (t--) {
    scanf("%s", s + 1);
    len = strlen(s + 1);
    int c1 = 0, c2 = 0;
    for (int i = 1; i < len; ++i) {
      if (s[i] == 'a' && s[i + 1] == 'b') ++c1;
      if (s[i] == 'b' && s[i + 1] == 'a') ++c2;
    }
    if (c1 != c2) {
      if (s[1] == 'a')
        s[1] = 'b';
      else
        s[1] = 'a';
    }
    printf("%s\n", s + 1);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
inline int read() {
  bool ok = 0;
  int res = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    ok = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  return ok ? ~res + 1 : res;
}
int t, n, m, k, x, y, len;
char s[110];
int main() {
  t = read();
  while (t--) {
    scanf("%s", s + 1);
    len = strlen(s + 1);
    int c1 = 0, c2 = 0;
    for (int i = 1; i < len; ++i) {
      if (s[i] == 'a' && s[i + 1] == 'b') ++c1;
      if (s[i] == 'b' && s[i + 1] == 'a') ++c2;
    }
    if (c1 != c2) {
      if (s[1] == 'a')
        s[1] = 'b';
      else
        s[1] = 'a';
    }
    printf("%s\n", s + 1);
  }
  return 0;
}

```