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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class GlobalRound7D2Solution2 {

	public static void solve() {
		PreCompute();
		int t = s.nextInt();
		while(t-- > 0) {
			String str = s.next();
			int n = str.length();
			int start = 0, end = n -1;
			StringBuilder ans = new StringBuilder();
			StringBuilder back = new StringBuilder();
			while(start < end && str.charAt(start) == str.charAt(end)) {
				ans.append(str.charAt(start));
				back.append(str.charAt(start));
				start++;
				end--;
			}
			back.reverse();
			if(start == end || start > end) {
				out.println(str);
				continue;
			}
			StringBuilder rem = new StringBuilder();
			for(int i = start; i <= end; i++) {
				rem.append(str.charAt(i));
			}
			int prefix = max_matching(rem.toString());
			int suffix = max_matching(rem.reverse().toString());
			if(prefix > suffix) {
				rem.reverse();
				for(int i = 0; i < prefix; i++) {
					ans.append(rem.charAt(i));
				}
			}else {
				for(int i = 0; i < suffix; i++) {
					ans.append(rem.charAt(i));
				}
			}
			ans.append(back);
			out.println(ans);
		}
	}
	
	//If tle comes we can reduce number of hashes to even two...
	public static int[] m = new int[] { (int) 1e9 + 7, (int) 982451653, (int) 2038074743 };
	public static int[] p = new int[] { (int) 3571, (int) 7919, (int) 541 };
	public static int max = (int) 1e6 + 10;// reduce it to 1e5 based on inputs
	public static long[][] pre = new long[2][max];
	public static int[][] hashes;
	public static int[][] shashes;
	public static void PreCompute() {
		//creates powers for each hash
		pre[0][0] = pre[1][0] = 1;
		for (int i = 0; i < 2; i++) {
			for (int j = 1; j < max; j++) {
				pre[i][j] = (pre[i][j - 1] * p[i]) % m[i];
			}
		}
	}
	
	public static int get(char c) {
		return (int)c - 'a' + 1;
	}
	
	public static void Prefix(String str) {
		hashes = new int[2][str.length()];
		for(int j = 0; j < 2; j++) {
			for(int i = 0; i < str.length(); i++) {
				if(i == 0) {
					hashes[j][0] = get(str.charAt(i));
				}else {
					long val = hashes[j][i-1];
					val += (get(str.charAt(i)) * (long)pre[j][i]);
					val %= m[j];
					hashes[j][i] = (int)val;
				}
			}
		}
	}
	
	public static void Suffix(String str) {
		shashes = new int[2][str.length()];
		int n = str.length();
		for(int j = 0; j < 2; j++) {
			for(int i = str.length() - 1; i >= 0; i--) {
				if(i == str.length() - 1) {
					shashes[j][0] = get(str.charAt(i));
				}else {
					long val = shashes[j][i+1];
					val += (get(str.charAt(i)) * (long)pre[j][n - i]);
					val %= m[j];
					shashes[j][i] = (int)val;
				}
			}
		}
	}
	
	public static int max_matching(String str) {
		Prefix(str);
		Suffix(str);
		int max = 1;
		int n = str.length();
		for(int i = 1; i < n; i++) {
			//Acting as middle character of pallindrome
			if((i&1)==1 && check(0, i/2, i/2+1, i)) {
				max = i + 1;
			}else if((i&1)==0 && check(0, i/2, i/2, i)) {
				max = i + 1;
			}
		}
		return max;
	}
	
	public static boolean check(int s1, int e1, int s2, int e2) {
		int n = hashes[0].length;
		for(int i = 0; i < 2; i++) {
			int hash1 = hashes[i][e1] - (s1 == 0 ? 0 : hashes[i][s1 - 1]);
			long hash_ = ((long)hash1 + (long)m[i])%m[i]; hash1 = (int)hash_;
//			System.out.println(e2);
			int hash2 = shashes[i][s2] - (e2 == n - 1 ? 0 : shashes[i][e2 + 1]);
			hash_ = ((long)hash2 + (long)m[i])%m[i]; hash2 = (int)hash_;
			long val = (long)hash1 * pre[i][n - e2];
			val%= m[i];
			if((int)val != hash2) {
				return false;
			}
		}
		return true;
	}
	
	public static void main(String[] args) {
		out = new PrintWriter(new BufferedOutputStream(System.out));
		s = new FastReader();
		solve();
		out.close();
	}

	public static FastReader s;
	public static PrintWriter out;

	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	
}
