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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.HashMap;
import java.util.Scanner;

import javax.print.attribute.HashAttributeSet;

public class d {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		//System.out.println(find2("cbb") );
		while (t-- > 0) {
			
			
			String as = s.next();
			if(as.length()==1) {
				System.out.println(as);
				continue;
			}
			int max = 0;
			String a = find(as);
			max = a.length();
			int n = as.length();
		 // System.out.println(a);

			String backpl = find2(as);
			
			if(backpl.length() > max) {
				a = backpl;
				max = backpl.length();
			}
			  
		//	System.out.println(a);

			
			String temp1 = "";
			String temp2="";
			int l = 0 ;
			int m =n-1;
			while(l<m &&as.charAt(l)==as.charAt(m) ) {
				l++;
				m--;
			}
			
			temp1 = as.substring(0 , l);
			temp2=as.substring(m+1);
			String temp3 = "";
			int max2 = 0 ;
		   String newas = as.substring(l , m+1);
			
			temp3 = find(newas);
			max2 = temp3.length();
					
		  backpl = find2(newas);
					if(max2 < backpl.length()) {
			
					temp3 = backpl ;}
					
					
			if(max > (temp1.length()+temp2.length()+temp3.length())) {
				System.out.println(a);
			}else {
				System.out.println(temp1+temp3+temp2);
			}
		}
		
		
	}

	

	static String find(String text) 
	{ 
		int N = text.length(); 
		if (N == 0) 
			return ""; 
		N = 2 * N + 1; // Position count 
		int[] L = new int[N + 1]; // LPS Length Array 
		L[0] = 0; 
		L[1] = 1; 
		int C = 1; // centerPosition 
		int R = 2; // centerRightPosition 
		int i = 0; // currentRightPosition 
		int iMirror; // currentLeftPosition 
		int maxLPSLength = 0; 
		int maxLPSCenterPosition = 0; 
		int start = -1; 
		int end = -1; 
		int diff = -1; 

		// Uncomment it to print LPS Length array 
		// printf("%d %d ", L[0], L[1]); 
		for (i = 2; i < N; i++) 
		{ 

			// get currentLeftPosition iMirror 
			// for currentRightPosition i 
			iMirror = 2 * C - i; 
			L[i] = 0; 
			diff = R - i; 

			// If currentRightPosition i is within 
			// centerRightPosition R 
			if (diff > 0) 
				L[i] = Math.min(L[iMirror], diff); 

			// Attempt to expand palindrome centered at 
			// currentRightPosition i. Here for odd positions, 
			// we compare characters and if match then 
			// increment LPS Length by ONE. If even position, 
			// we just increment LPS by ONE without 
			// any character comparison 
			while (((i + L[i]) + 1 < N && (i - L[i]) > 0) && 
							(((i + L[i] + 1) % 2 == 0) || 
						(text.charAt((i + L[i] + 1) / 2) == 
						text.charAt((i - L[i] - 1) / 2)))) 
			{ 
				L[i]++; 
			} 

			if (L[i] > maxLPSLength && (i - L[i])/2==0) // Track maxLPSLength 
			{ 
				maxLPSLength = L[i]; 
				maxLPSCenterPosition = i; 
			} 

			// If palindrome centered at currentRightPosition i 
			// expand beyond centerRightPosition R, 
			// adjust centerPosition C based on expanded palindrome. 
			if (i + L[i] > R) 
			{ 
				C = i; 
				R = i + L[i]; 
			} 

			// Uncomment it to print LPS Length array 
			// printf("%d ", L[i]); 
		} 

		start = (maxLPSCenterPosition - maxLPSLength) / 2; 
		end = start + maxLPSLength - 1; 
		String ans = "";
		//System.out.printf("LPS of string is %s : ", text); 
		for (i = start; i <= end; i++) 
			ans+=text.charAt(i);
		if(ans.equals("")) {
			return text.charAt(0)+"";
		}
		return ans;
		
	} 


	static String find2(String text) 
	{ //System.out.println(text);
		int N = text.length(); 
		if (N == 0) 
			return ""; 
		N = 2 * N + 1; // Position count 
		int[] L = new int[N + 1]; // LPS Length Array 
		L[0] = 0; 
		L[1] = 1; 
		int C = 1; // centerPosition 
		int R = 2; // centerRightPosition 
		int i = 0; // currentRightPosition 
		int iMirror; // currentLeftPosition 
		int maxLPSLength = 0; 
		int maxLPSCenterPosition = 0; 
		int start = -1; 
		int end = -1; 
		int diff = -1; 

		// Uncomment it to print LPS Length array 
		// printf("%d %d ", L[0], L[1]); 
		for (i = 2; i < N; i++) 
		{ 

			// get currentLeftPosition iMirror 
			// for currentRightPosition i 
			iMirror = 2 * C - i; 
			L[i] = 0; 
			diff = R - i; 

			// If currentRightPosition i is within 
			// centerRightPosition R 
			if (diff > 0) 
				L[i] = Math.min(L[iMirror], diff); 

			// Attempt to expand palindrome centered at 
			// currentRightPosition i. Here for odd positions, 
			// we compare characters and if match then 
			// increment LPS Length by ONE. If even position, 
			// we just increment LPS by ONE without 
			// any character comparison 
			while (((i + L[i]) + 1 < N && (i - L[i]) > 0) && 
							(((i + L[i] + 1) % 2 == 0) || 
						(text.charAt((i + L[i] + 1) / 2) == 
						text.charAt((i - L[i] - 1) / 2)))) 
			{ 
				L[i]++; 
			} 

			if (L[i] > maxLPSLength && ((i - L[i])/2 + L[i] -1) ==text.length()-1) // Track maxLPSLength 
			{ 
				maxLPSLength = L[i]; 
				maxLPSCenterPosition = i; 
			} 

			// If palindrome centered at currentRightPosition i 
			// expand beyond centerRightPosition R, 
			// adjust centerPosition C based on expanded palindrome. 
			if (i + L[i] > R) 
			{ 
				C = i; 
				R = i + L[i]; 
			} 

			// Uncomment it to print LPS Length array 
			// printf("%d ", L[i]); 
		} 

		start = (maxLPSCenterPosition - maxLPSLength) / 2; 
		end = start + maxLPSLength - 1; 
		String ans = "";
		//System.out.printf("LPS of string is %s : ", text); 
		for (i = start; i <= end; i++) 
			ans+=text.charAt(i);
		
		
		return ans;
		
	} 

}