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.math.*;
import java.io.*;
import java.util.*;

public class Solution {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		int t = sc.nextInt();
		while (t-->0) {
			String s = sc.nextLine();
			int n = s.length();
			int i=0,j=n-1;
			while(i<j){
			    if(s.charAt(i)==s.charAt(j)){
			        i++;
			        j--;
			        continue;
			    }
			    break;
			}
			StringBuilder rev = new StringBuilder();
			String p = s.substring(i,j+1);
			for (int i1=p.length()-1;i1>=0 ;i1-- ) rev.append(p.charAt(i1));
			String rev1 = rev.toString();
			String p1 = getit(p,rev1);
			String p2 = getit(rev1,p);
			if(p1.length()>=p2.length()){
			    System.out.println(s.substring(0,i)+""+p1+""+s.substring(j+1));
			}
			else{
			    System.out.println(s.substring(0,i)+""+p2+""+s.substring(j+1));
			}
		}
	}
	static String getit(String s, String b){
	    StringBuilder temp=new StringBuilder();
        temp.append(s);
        temp.append("#");
        temp.append(b);
	    int n = temp.length();
	    int[] pi = new int[n];
	    computeLPSArray(temp.toString(),n,pi);
	    return s.substring(0,pi[n-1]);
	}
	static void computeLPSArray(String pat, int M, int lps[]) 
    { 
        // length of the previous longest prefix suffix 
        int len = 0; 
        int i = 1; 
        lps[0] = 0; // lps[0] is always 0 
  
        // the loop calculates lps[i] for i = 1 to M-1 
        while (i < M) { 
            if (pat.charAt(i) == pat.charAt(len)) { 
                len++; 
                lps[i] = len; 
                i++; 
            } 
            else // (pat[i] != pat[len]) 
            { 
                // This is tricky. Consider the example. 
                // AAACAAAA and i = 7. The idea is similar 
                // to search step. 
                if (len != 0) { 
                    len = lps[len - 1]; 
  
                    // Also, note that we do not increment 
                    // i here 
                } 
                else // if (len == 0) 
                { 
                    lps[i] = len; 
                    i++; 
                } 
            } 
        } 
    } 
	static class Query {
		int x1, x2, y, id;
		public Query(int c, int d, int e, int a){
			x1 = c;
			x2 = d;
			y = e;
			id = a;
		}
	}
	static class Compare1 implements Comparator<Query>{
		@Override
		public int compare(Query q1, Query q2){
			if(q1.x2==q2.x2) return q1.x1-q2.x1;
			return q1.x2-q2.x2;
		}
	}
	static class Pair implements Comparable<Pair>{
		int a, b;
		public Pair(int c, int d){
			a = c;
			b = d;
		}
		public int compareTo(Pair p){
		    return a-p.a;
		}
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}
		FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}
		String nextLine() {
			try {
				return br.readLine();
			} catch (IOException e) {
				return null;
			}
		}
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}