Problem: This is the easy 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 ≤ 1000), 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 5000.

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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

/*
1
abcdfdcecba

1
codeforces

1
ababacaba
1
sasdfdsababasdfdss

1
abac
 */
public class D2 {

	static HarmeyerHash[] prefixes, suffixes;
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int T=fs.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		for (int tt=0; tt<T; tt++) {
			char[] a=fs.next().toCharArray();
			int n=a.length;
			prefixes=HarmeyerHash.getPrefixHashes(a);
			char[] rev=new char[n];
			for (int i=0; i<n; i++) rev[i]=a[n-1-i];
			suffixes=HarmeyerHash.getPrefixHashes(rev);
			
			if (isPal(n, 0, n)) {
				out.println(a);
			}
			else {
				for (int i=0; i<n; i++) {
					if (a[i]!=a[n-1-i]) {
						int frontRemove=removeCountFront(i, n);
						int backRemove=removeBackFront(n-i-1, n);
//						System.out.println("Broke at "+i+" "+frontRemove+" "+backRemove);
						ArrayList<Character> ans=new ArrayList<>();
						if (frontRemove<=backRemove) {
							for (int j=0; j<n; j++) {
								if (j<i || j>=i+frontRemove) ans.add(a[j]);
							}
						}
						else {
							for (int j=0; j<n; j++) {
								if (j>n-1-i|| j<=n-1-i-backRemove) ans.add(a[j]);
							}
						}
						for (char c:ans) out.print(c);
						out.println();
						break;
					}
				}
			}
		}
		out.close();
	}
	
	static int removeCountFront(int firstIssue, int n) {
		int farIndex=firstIssue;
		while (!isPal(firstIssue, n-farIndex, n)) {
			farIndex++;
		}
		return farIndex-firstIssue;
	}
	static int removeBackFront(int lastIssue, int n) {
		int frontIndex=lastIssue;
		while (!isPal(frontIndex+1, n-1-lastIssue, n)) {
			frontIndex--;
		}
		return lastIssue-frontIndex;
	}
	
	static boolean isPal(int preLen, int postLen, int n) {
//		System.out.println("Trying isPal "+preLen+" "+postLen);
		return get(preLen, postLen, prefixes, n).compareTo(get(postLen, preLen, suffixes, n))==0;
	}
	
	static HarmeyerHash get(int preLen, int postLen, HarmeyerHash[] p, int n) {
		HarmeyerHash ret=HarmeyerHash.substringHash(p, 0, preLen);
		ret.append(HarmeyerHash.substringHash(p, n-postLen, n));
		return ret;
	}
	
	static class HarmeyerHash implements Comparable<HarmeyerHash> {
		static final long m1=8675309, m2=1_000_000_007;
		long v1=0, v2=0; int l=0;	
		static final long s1=257, s2=619;
		static long[] s1Pow, s2Pow;
		static boolean precomped=false;
		
		void add(char o) {
			v1=(v1*s1+o)%m1;
			v2=(v2*s2+o)%m2;
			l++;
		}
		
		public int compareTo(HarmeyerHash o) {
			if (v1!=o.v1)
				return Long.compare(v1, o.v1);
				return Long.compare(v2, o.v2);
		}
		
		public boolean equals(Object o) {
			return compareTo((HarmeyerHash)o)==0;
		}

		public int hashCode() {
			return (int)v1;
		}
		
		public HarmeyerHash clone() {
			HarmeyerHash toReturn=new HarmeyerHash();
			toReturn.v1=v1;
			toReturn.v2=v2;
			toReturn.l=l;
			return toReturn;
		}

		static void precomp() {
			if (precomped) return;
			precomped=true;
			s1Pow=new long[1000_002];
			s2Pow=new long[1000_002];
			s1Pow[0]=s2Pow[0]=1;
			for (int i=1; i<s1Pow.length; i++)
				s1Pow[i]=(s1Pow[i-1]*s1)%m1;
			for (int i=1; i<s2Pow.length; i++)
				s2Pow[i]=(s2Pow[i-1]*s2)%m2;
		}

		//need fastPow if o can be longer than 10^6
		void append(HarmeyerHash o) {
			precomp();
			v1=(v1*s1Pow[o.l]+o.v1)%m1;
			v2=(v2*s2Pow[o.l]+o.v2)%m2;
			l+=o.l;
		}
		
		public static HarmeyerHash[] getPrefixHashes(char[] word) {
			precomp();
			int n=word.length;
			HarmeyerHash[] toReturn=new HarmeyerHash[n+1];
			toReturn[0]=new HarmeyerHash();
			for (int i=1; i<=n; i++) {
				toReturn[i]=toReturn[i-1].clone();
				toReturn[i].add(word[i-1]);
			}
			return toReturn;
		}
		
		//inclusive, exclusive
		public static HarmeyerHash substringHash(HarmeyerHash[] prefixHashes, int from, int to) {
			if (from==to)
				return new HarmeyerHash();
			HarmeyerHash old=prefixHashes[to].clone(), toSub=prefixHashes[from];
			int diff=to-from;
			old.v1=(old.v1-(toSub.v1*s1Pow[diff])%m1+m1)%m1;
			old.v2=(old.v2-(toSub.v2*s2Pow[diff])%m2+m2)%m2;
			old.l=to-from;
			return old;
		}
		
	}



	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		boolean hasNext() {
			String next=null;
			try {
				next = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (next==null) {
				return false;
			}
			st=new StringTokenizer(next);
			return true;
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}

}