#include <bits/stdc++.h>
using namespace std;
void ha() { printf("YES\n"); }
void na() { printf("NO\n"); }
long long power(long long a, long long b) {
  long long c, d;
  if (b == 1) {
    return a % 1000000007;
  } else if (b % 2 == 0) {
    c = power(a, b / 2);
    return (c * c) % 1000000007;
  } else {
    c = power(a, b / 2);
    d = (c * c) % 1000000007;
    d = (d * a) % 1000000007;
    return d;
  }
}
int absolute(int a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
long long absolute(long long a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    string a;
    cin >> a;
    int n = a.length();
    if (a[0] != a[n - 1]) {
      a[0] = a[n - 1];
      cout << a << endl;
    } else {
      cout << a << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
void ha() { printf("YES\n"); }
void na() { printf("NO\n"); }
long long power(long long a, long long b) {
  long long c, d;
  if (b == 1) {
    return a % 1000000007;
  } else if (b % 2 == 0) {
    c = power(a, b / 2);
    return (c * c) % 1000000007;
  } else {
    c = power(a, b / 2);
    d = (c * c) % 1000000007;
    d = (d * a) % 1000000007;
    return d;
  }
}
int absolute(int a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
long long absolute(long long a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    string a;
    cin >> a;
    int n = a.length();
    if (a[0] != a[n - 1]) {
      a[0] = a[n - 1];
      cout << a << endl;
    } else {
      cout << a << endl;
    }
  }
  return 0;
}

```