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

    #from _typeshed import SupportsKeysAndGetItem
import sys
from collections import deque
import heapq
import bisect
import copy
# sys.stdin=open("input.txt","r");
# sys.stdout=open("output.txt","w")
####### GLOBAL ###############
MOD=1000000007
NO=lambda:print("NO")
YES=lambda:print("YES")
_1=lambda:print(-1)
ari=lambda:[int(_) for _ in input().split()]
cin=lambda:int(input())
cis=lambda:input()
show=lambda x: print(x)
fast= lambda:sys.stdin.readline()

########### END #########
######
test_case=1
test_case=int(input())
######
sys.setrecursionlimit(10**6)
    
def ans():
    s=cis()
    abcount=0
    bacount=0
    n=len(s)
    s=list(s)
    def bruteforce(news):
        abcount=0
        nonlocal n
        bacount=0
        for i in range(n-1):
            if s[i]=="a" and s[i+1]=="b":
                abcount+=1
            if s[i]=="b" and s[i+1]=="a":
                bacount+=1
        # print(abcount,bacount)
        if abcount==bacount:
            return "".join(news)
        return False
    ans=bruteforce(s)
    if bruteforce(s)!=False:
        print(ans)
        return 
        
    for i in range(n):
        val=s[i]
        if s[i]=="a":
            s[i]="b"
            ans=bruteforce(s)
            if ans!=False:
                print(ans)
                return
            s[i]="a"
        else:
            s[i]="a"
            ans=bruteforce(s)
            if ans!=False:
                print(ans)
                return
            s[i]="b"
    print(ans)
            
            
    # print("b"*len(a))
    return
for _ in range(test_case):
    ans()
    