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".
import java.io.*;
import java.util.*;
public class Main {
	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 String LongestPalindromicPrefix(String str)
	{
	  
	    // 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 static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
		t=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		for(int tt=0;tt<t;tt++) {
			String a=f.next();
			int start=0;
			int end=a.length()-1;
			StringBuffer first=new StringBuffer();
			StringBuffer last=new StringBuffer();
			while(start<=end) {
				if(a.charAt(start)==a.charAt(end) && start<end) {
					first.append(a.charAt(start));
					last.append(a.charAt(end));
					start++;
					end--;
				}
				else if(a.charAt(start)==a.charAt(end) && start==end) {
					first.append(a.charAt(start));
					start++;
					end--;
				}
				else {
					break;
				}
			}
			last.reverse();
			String mid="";
			if(start<end) {
				String mida=LongestPalindromicPrefix(a.substring(start,end+1));
				String midb=LongestPalindromicPrefix(reverse(a.substring(start,end+1)));
//				System.out.println(rev(a.substring(start,end)));
//				System.out.println(midb);
				if(midb.length()>mida.length()) {
					mid=reverse(midb);
				}
				else {
					mid=mida;
				}
			}
			StringBuffer ans=new StringBuffer();
			ans.append(first);
			ans.append(mid);
			ans.append(last);
			out.println(first+mid+last);
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	
//Some things to notice
//Check for the overflow
//Binary Search
//Bitmask