This is the easy version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.

You are given a string s, consisting of lowercase English letters. Find the longest string, t, which satisfies the following conditions: 

  * The length of t does not exceed the length of s. 
  * t is a palindrome. 
  * There exists two strings a and b (possibly empty), such that t = a + b ( "+" represents concatenation), and a is prefix of s while b is suffix of s. 

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000), the number of test cases. The next t lines each describe a test case.

Each test case is a non-empty string s, consisting of lowercase English letters.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 5000.

Output

For each test case, print the longest string which satisfies the conditions described above. If there exists multiple possible solutions, print any of them.

Example

Input


5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba


Output


a
abcdfdcba
xyzyx
c
abba

Note

In the first test, the string s = "a" satisfies all conditions.

In the second test, the string "abcdfdcba" satisfies all conditions, because:

  * Its length is 9, which does not exceed the length of the string s, which equals 11. 
  * It is a palindrome. 
  * "abcdfdcba" = "abcdfdc" + "ba", and "abcdfdc" is a prefix of s while "ba" is a suffix of s. 



It can be proven that there does not exist a longer string which satisfies the conditions.

In the fourth test, the string "c" is correct, because "c" = "c" + "" and a or b can be empty. The other possible solution for this test is "s".
mod=998244353
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**27)
import sys
#sys.setrecursionlimit(10**6)
#fact=[1]
#for i in range(1,100001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
#from random import randint as rn
#from Queue import Queue as Q
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%mod))%mod
    return t
def ain():                           #takes array as input
    return list(map(int,stdin.readline().split()))
def sin():
    return input().strip()
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
"""*******************************************************"""
def main():
    for _ in range(int(input())):
        s=sin()
        n=len(s)
        d=0
        for i in range(n/2):
            if(s[i]==s[n-i-1]):
                d+=1
            else:
                break
        ans1=0
        for i in range(n-d,d,-1):
            fl=0
            for j in range(d,i):
                if(s[j]!=s[i-1+d-j]):
                    fl=1
                    break
            if(fl==0):
                ans1=i-d
                break
        ans2=0
        for i in range(d,n-d):
            fl=0
            for j in range(i,n-d):
                if(s[j]!=s[n-d-1-j+i]):
                    fl=1
                    break
            if(fl==0):
                ans2=n-d-i
                break
        if(ans1>ans2):
            print s[:d+ans1]+s[n-d:]
        else:
            print s[:d]+s[n-d-ans2:]

####### Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

if __name__ == '__main__':
   main()
#threading.Thread(target=main).start()
