This is the hard 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 ≤ 10^5), 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 10^6.

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".
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by enraged(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import heapq
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt
from collections import defaultdict as dd, deque, Counter as c
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect

# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353


def data(): return sys.stdin.readline().strip()


def out(var): sys.stdout.write(str(var))


def l(): return list(sp())


def sl(): return list(ssp())


def sp(): return map(int, data().split())


def ssp(): return map(str, data().split())


def l1d(n, val=0): return [val for i in range(n)]


def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


def post_process(string, ml, c):
    result = []
    for i in string[c-ml:c+ml+1]:
        if i == '#':
            continue
        result.append(i)
    return ''.join(result)


def manacher(string):
    arr = ['#']
    for i in string:
        arr.append(i)
        arr.append('#')
    left = l1d(len(arr))
    r, c = 0, 0
    for i in range(len(arr)):
        mirror = 2 * c - i
        if r > i:
            left[i] = min(r - i, left[mirror])
        try:
            while arr[i + 1 + left[i]] == arr[i - 1 - left[i]]:
                left[i] += 1
        except:
            pass
        if i + left[i] > r:
            c = i
            r = i + left[i]
    maxlength = 0
    center = 0
    for i in range(len(left)):
        if left[i] > maxlength and (i == left[i] or (i + left[i]) >= len(left)-1):
            maxlength = left[i]
            center = i
    return post_process(arr, maxlength, center)


for _ in range(int(data())):
    s = data()
    if s[::-1] == s:
        out(s + "\n")
        continue
    i, j = 0, len(s) - 1
    while i < j:
        if s[i] == s[j]:
            i += 1
            j -= 1
        else:
            break
    if i == 0:
        out(manacher(s)+"\n")
    else:
        out(s[:i]+manacher(s[i:j+1])+s[j+1:]+"\n")
