human
Write me a Python code to print the resulting string s with \operatorname{AB}(s) = \operatorname{BA}(s) you'll get making the minimum number of steps given the following:
Can you solve the below in PYTHON3?
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.
assistant
#Here is Python code to print the resulting string s with \operatorname{AB}(s) = \operatorname{BA}(s) you'll get making the minimum number of steps.

def main():
    alpha = 'abcdefghijklmnopqrstuvwxyz'
    ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    inf = 1e17
    # mod = 998244353
    mod = 10 ** 9 + 7

    # Max = 10 ** 6
    # primes = []
    # prime = [True for i in range(Max + 1)]
    # p = 2
    # while (p * p <= Max + 1):
    #
    #     # If prime[p] is not
    #     # changed, then it is a prime
    #     if (prime[p] == True):
    #
    #         # Update all multiples of p
    #         for i in range(p * p, Max + 1, p):
    #             prime[i] = False
    #     p += 1
    #
    # for p in range(2, Max + 1):
    #     if prime[p]:
    #         primes.append(p)

    # print(len(primes))

    def factorial(n):
        f = 1

        for i in range(1, n + 1):
            f = (f * i) % mod  # Now f never can
        # exceed 10^9+7

        return f

    def factorial_by(n, by):
        f = 1

        for i in range(1, n + 1):
            if i == by:
                continue
            f = (f * i) % mod  # Now f never can
        # exceed 10^9+7

        return f

    def ncr(n, r):
        # initialize numerator
        # and denominator
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % mod
            den = (den * (i + 1)) % mod
        return (num * pow(den,
                          mod - 2, mod)) % mod

    def primeFactors(num):

        pf = []
        while num % 2 == 0:
            pf.append(2)
            num = num // 2

        # n must be odd at this point
        # so a skip of 2 ( i = i + 2) can be used
        for i in range(3, int(math.sqrt(num)) + 1, 2):

            # while i divides n , print i and divide n
            while num % i == 0:
                pf.append(i)
                num = num // i

        # Condition if n is a prime
        # number greater than 2
        if num > 2:
            pf.append(num)

        return pf

    class Node(object):
        def __init__(self, name, d):
            self.d = d
            self.name = name

        def __repr__(self):
            pass

   ...