Problem: 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".

Solution (in PYTHON3):

Solution:
def getLPSFromIEq0_len(text):
    N = len(text)
    if N == 0:
        return
    N = 2*N+1    # Position count
    L = [0] * N
    L[0] = 0
    L[1] = 1
    C = 1     # centerPosition
    R = 2     # centerRightPosition
    i = 0    # currentRightPosition
    iMirror = 0     # currentLeftPosition
    maxLPSLength = 0
    maxLPSCenterPosition = 0
    start = -1
    end = -1
    diff = -1
    
    lpsFromIEq0Len=1
  
    # Uncomment it to print LPS Length array
    # printf("%d %d ", L[0], L[1]);
    for i in range(2,N):
      
        # get currentLeftPosition iMirror for currentRightPosition i
        iMirror = 2*C-i
        L[i] = 0
        diff = R - i
        # If currentRightPosition i is within centerRightPosition R
        if diff > 0:
            L[i] = min(L[iMirror], diff)
  
        # Attempt to expand palindrome centered at currentRightPosition i
        # Here for odd positions, we compare characters and
        # if match then increment LPS Length by ONE
        # If even position, we just increment LPS by ONE without
        # any character comparison
        try:
            while ((i+L[i]) < N and (i-L[i]) > 0) and \
                    (((i+L[i]+1) % 2 == 0) or \
                    (text[(i+L[i]+1)//2] == text[(i-L[i]-1)//2])):
                L[i]+=1
        except Exception as e:
            pass
  
        if L[i] > maxLPSLength:        # Track maxLPSLength
            maxLPSLength = L[i]
            maxLPSCenterPosition = i
        
        #My injected code to find LPS starting from i==0
        if i==L[i]:
            lpsFromIEq0Len=max(lpsFromIEq0Len,L[i])
  
        # If palindrome centered at currentRightPosition i
        # expand beyond centerRightPosition R,
        # adjust centerPosition C based on expanded palindrome.
        if i + L[i] > R:
            C = i
            R = i + L[i]
  
    # Uncomment it to print LPS Length array
    # printf("%d ", L[i]);
    # start = (maxLPSCenterPosition - maxLPSLength) // 2
    # end = start + maxLPSLength - 1
    # print ("LPS of string is " + text + " : ")
    # print (text[start:end+1])
    return lpsFromIEq0Len
  

def main():
    
    q=int(input())
    allans=[]
    for _ in range(q):
        s=input()
        i=0
        j=len(s)-1
        while i<=j:
            if s[i]==s[j]:
                i+=1
                j-=1
            else:
                break
        if i>j:
            allans.append(s)
            continue
        middleS=s[i:j+1]
        l1=getLPSFromIEq0_len(middleS)
        l2=getLPSFromIEq0_len(middleS[::-1])
        if l1>=l2:
            innerPalindrome=middleS[:l1]
        else:
            innerPalindrome=middleS[-l2:]
        ans=s[:i]+innerPalindrome+s[j+1:]
        allans.append(ans)
    multiLineArrayPrint(allans)
    
    return

import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]

def makeArr(defaultVal,dimensionArr): # eg. makeArr(0,[n,m])
    dv=defaultVal;da=dimensionArr
    if len(da)==1:return [dv for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(x,y):
    print('? {} {}'.format(x,y))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(ans):
    print('! {}'.format(ans))
    sys.stdout.flush()
 
inf=float('inf')
MOD=10**9+7
# MOD=998244353
 
 
for _abc in range(1):
    main()