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 JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

 public class Solution implements Runnable {
    
    
	 static String LongestPalindromicPrefix(String str) throws Exception
	 { 
	   
	     // Create temporary String 
	     String temp = str + '?'; 
	   
	     // Reverse the String str 
	   
	    str=reverse(str);
	   
	     // Append String str to temp 
	     temp += str; 
	   
	     // Find the length of String temp 
	     int n = temp.length(); 
	   
	     // lps[] array for String temp 
	     int []lps = new int[n]; 
	   
	     // Intialise every value with zero 
	     Arrays.fill(lps, 0); 
	   
	     // Iterate the String temp 
	     for(int i = 1; i < n; i++) 
	     { 
	           
	        // Length of longest prefix 
	        // till less than i 
	        int len = lps[i - 1]; 
	          
	        // Calculate length for i+1 
	        while (len > 0 && temp.charAt(len) !=  
	                          temp.charAt(i))  
	        { 
	            len = lps[len - 1]; 
	        } 
	          
	        // If character at current index 
	        // len are same then increament 
	        // length by 1 
	        if (temp.charAt(i) == temp.charAt(len)) 
	        { 
	            len++; 
	        } 
	          
	        // Update the length at current 
	        // index to len 
	        lps[i] = len; 
	     } 
	   
	     // Print the palindromic String  
	     // of max_len 
	     return (temp.substring(0, lps[n - 1])); 
	 } 
    
    public void solve() throws Exception {
    	int t=sc.nextInt();
    		while(t-->0)
    		{
    			String s=sc.nextToken();
    			int len=s.length();
    			int count=0;
    			int n=len;
    			int max=Integer.MAX_VALUE;
    			int i=0;
    			for(i=0;i<n;i++)
    			{
    				if(s.charAt(i)==s.charAt(len-i-1))
    					count++;
    				else break;
    				if(i>=len-i-1)
    					break;
    			}
    			int first_check=i;
    			int second_check=n-i-1;
    			String send=s.substring(first_check,second_check+1);
    					String send2="";
    			
    			String x=LongestPalindromicPrefix(send);
    			send2=reverse(send);
    			String y=LongestPalindromicPrefix(send2);
    			
    		System.out.println(s.substring(0,first_check)+(x.length()>y.length()?x:y) +s.substring(second_check+1));
    			
    		}
    }
    static String reverse(String input) 
    { 
        char[] a = input.toCharArray(); 
        int l, r = a.length - 1; 
          
        for(l = 0; l < r; l++, r--) 
        { 
           char temp = a[l]; 
           a[l] = a[r]; 
           a[r] = temp; 
        } 
        return String.valueOf(a); 
    } 
    
    
    static Throwable uncaught;

    BufferedReader in;
    FastScanner sc;
    PrintWriter out;

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            sc = new FastScanner(in);
            solve();
        } catch (Throwable uncaught) {
            Solution.uncaught = uncaught;
        } finally {
            out.close();
        }
    }

    public static void main(String[] args) throws Throwable {
        Thread thread = new Thread(null, new Solution(), "", (1 << 26));
        thread.start();
        thread.join();
        if (Solution.uncaught != null) {
            throw Solution.uncaught;
        }
    }

}

class FastScanner {

    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }

    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }

}