Problem: 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".

Solution (in PYTHON):

Solution:
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR(): return list(input())

    def INT(): return int(input())

    def MAP(): return map(int, input().split())

    def LIST(): return list(map(int, input().split()))

    def list2d(a, b, c): return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):   return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    # cook here:
    def KMP(s):
        n = len(s)
        lps = [0] * n  # lps[0] is always 0

        # length of the previous
        # longest prefix suffix
        l = 0

        # the loop calculates lps[i]
        # for i = 1 to n-1
        i = 1
        while (i < n):
            if (s[i] == s[l]):
                l = l + 1
                lps[i] = l
                i = i + 1

            else:

                # (pat[i] != pat[len])
                # This is tricky. Consider
                # the example. AAACAAAA
                # and i = 7. The idea is
                # similar to search step.
                if (l != 0):
                    l = lps[l - 1]

                    # Also, note that we do
                    # not increment i here

                else:

                    # if (len == 0)
                    lps[i] = 0
                    i = i + 1

        res = lps[n - 1]

        # Since we are looking for
        # non overlapping parts.
        if (res > n / 2):
            return n // 2
        else:
            return res

    def find_largest_pal_pref(st):
        temp = st + str('?') + st[::-1]
        return KMP(temp)

    t = INT()

    while (t != 0):
        s = input()

        n = len(s)

        i = 0
        j = n - 1
        ans = 0
        while (i < j):
            if (s[i] == s[j]):
                ans += 1
            else:
                break
            i += 1
            j -= 1
        final = s[:i]
        # print(final,i,j)
        if (j - i + 1 > 0):
            rem = s[i:j + 1]
            # print(rem)
            temp = find_largest_pal_pref(rem)
            rem = rem[::-1]
            temp1 = find_largest_pal_pref(rem)

            if (temp > temp1):
                rem = rem[::-1]
                final += rem[:temp]
            else:
                # rem = rem[::-1]
                final += rem[:temp1]

        final += s[j + 1:]
        # if(n==1):
        #     final = s
        print(final)

        t -= 1
######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

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')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()