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 PYTHON3):

Solution:
import io,os
from collections import deque
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
RL = lambda : list(map(int, input().split()))



# Python3 program to find the longest  
# palindrome in a string formed by  
# concatenating its prefix and suffix 
   
# Function to check whether 
# the string is a palindrome 
def isPalindrome(r): 
  
    p = r 
   
    # Reverse the string to 
    # compare with the 
    # original string 
    p = "".join(reversed(p)) 
   
    # Check if both are same 
    return (r == p) 
   
# Function to find the longest 
# palindrome in a string formed by 
# concatenating its prefix and suffix 
def PrefixSuffixPalindrome(st): 
  
    # Length of the string 
    n = len(st) 
    length = 0
   
    # Finding the length upto which 
    # the suffix and prefix forms a 
    # palindrome together 
    for i in range( n // 2): 
        if (st[i] != st[n - i - 1]): 
            length = i 
            break
   
    # Check whether the string 
    # has prefix and suffix substrings 
    # which are palindromes. 
    prefix = "" 
    suffix = "" 
    midPal = "" 
   
    # Removing the suffix and prefix 
    # substrings which already forms 
    # a palindrome and storing them 
    # in separate strings 
    prefix = st[:length] 
    suffix = st[n - length:] 
    st = st[length: n - 2 * length+length] 
  
    # Check all prefix substrings 
    # in the remaining string str 
    for i in range(1,len(st)+1): 
        y = st[0: i] 
   
        # Check if the prefix substring 
        # is a palindrome 
        if (isPalindrome(y)): 
   
            # If the prefix substring 
            # is a palindrome then check 
            # if it is of maximum length 
            # so far 
            if (len(midPal) < len(y)): 
                midPal = y 
   
    # Check all the suffix substrings 
    # in the remaining string str 
    for i in range(1,len(st)+1): 
        y = st[len(st)-i:] 
        #print(y)
        # Check if the suffix substring 
        # is a palindrome 
        if (isPalindrome(y)): 
   
            # If the suffix substring 
            # is a palindrome then check 
            # if it is of maximum length 
            # so far 
            if (len(midPal) < len(y)): 
                midPal = y 
   
    # Combining all the thee parts 
    # of the answer 
    answer = prefix + midPal + suffix 
   
    return answer 
   
# Driver code 
if __name__ == "__main__": 
    T = int(input())
    for i in range(T):
        st = input()
        print(PrefixSuffixPalindrome(st))