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:
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		(⠀ ﾉ⌒ Y ⌒ヽ-く __／
		| _⠀｡ノ| ノ｡ |/
		(⠀ー '_人`ー ﾉ
		⠀|\ ￣ _人'彡ﾉ
		⠀ )\⠀⠀ ｡⠀⠀ /
		⠀⠀(\⠀ #⠀ /
		⠀/⠀⠀⠀/ὣ====================D-
		/⠀⠀⠀/⠀ \ \⠀⠀\
		( (⠀)⠀⠀⠀⠀ ) ).⠀)
		(⠀⠀)⠀⠀⠀⠀⠀( | /
		|⠀ /⠀⠀⠀⠀⠀⠀ | /
		[_] ⠀⠀⠀⠀⠀[___]                     */
	// Main Code at the Bottom
	import java.util.*;
	import java.lang.*;
	import java.io.*;
	import java.math.BigInteger; 
	public class Main {
		//Fast IO class
	    static class FastReader {
	        BufferedReader br; 
	        StringTokenizer st; 
	        public FastReader() {
	        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
	        	if(!env) {
	        		try {
						br=new BufferedReader(new FileReader("src\\input.txt"));
					} catch (FileNotFoundException e) {
						e.printStackTrace();
					}
	        	}
	        	else br = new BufferedReader(new InputStreamReader(System.in)); 
	        } 
	        String next() {
	            while (st == null || !st.hasMoreElements()) {
	                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()); 
	        } 
	        String nextLine() {
	            String str = ""; 
	            try {
	                str = br.readLine(); 
	            } 
	            catch (IOException e) {
	                e.printStackTrace(); 
	            } 
	            return str; 
	        } 
	    }     
	    static long MOD=1000000000+7;
	    //Euclidean Algorithm
	    static long gcd(long A,long B){
	    	return (B==0)?A:gcd(B,A%B);
	    }
	    //Modular Exponentiation
	    static long fastExpo(long x,long n){
	        if(n==0) return 1;
	        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
	        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
	    }
	    //Modular Inverse
	    static long inverse(long x) {
	    	return fastExpo(x,MOD-2);
	    }
	    //Prime Number Algorithm
	    static boolean isPrime(long n){
	        if(n<=1) return false;
	        if(n<=3) return true;
	        if(n%2==0 || n%3==0) return false;
	        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
	        return true;
	    }
	    //Reverse an array
	    static void reverse(int arr[],int l,int r){
	    	while(l<r) {
	    		int tmp=arr[l];
	    		arr[l++]=arr[r];
	    		arr[r--]=tmp;
	    	}
	    }
	    //Print array
	    static void print1d(int arr[]) {
	    	out.println(Arrays.toString(arr));
	    }
	    static void print2d(long arr[][]) {
	    	for(long a[]: arr) out.println(Arrays.toString(a));
	    }
	    // Pair
	    static class pair{
	    	int x,y;
	    	pair(int a,int b){
	    		this.x=a;
	    		this.y=b;
	    	}
	    	public boolean equals(Object obj) {
	    		if(obj == null || obj.getClass()!= this.getClass()) return false;
	            pair p = (pair) obj;
	            return (this.x==p.x && this.y==p.y);
	        }
	    	public int hashCode() {
	            return Objects.hash(x,y);
	        }
	    }
	    static FastReader sc=new FastReader();
	    static PrintWriter out=new PrintWriter(System.out);  
	    //Main function(The main code starts from here
	    public static void main (String[] args) throws java.lang.Exception {
	    	int test;
	    	//test=1;
	    	test=sc.nextInt();
	    	while(test-->0){
	    		char a[]=sc.nextLine().toCharArray();
	    		int n=a.length,l=0,r=n-1;
	    		while(l<r && a[l]==a[r]) {
	    			l++;
	    			r--;
	    		}
	    		if(l==r || r<l) {
	    			for(char c: a) out.print(c);
	    			out.println();
	    			continue;
	    		}
	    		int ans=l+n-r-1;
	    		StringBuilder left=new StringBuilder(),right=new StringBuilder(),tmp=new StringBuilder();
	    		for(int i=0;i<l;i++) left.append(a[i]);
	    		for(int i=r+1;i<n;i++) right.append(a[i]);
	    		int dp[][]=new int[n][n],cnt=0;
	    		for(int i=r;i>=l;i--) {
	    			for(int j=i;j<=r;j++) {
	    				if(a[i]==a[j]) {
	    					if(j-i<=2) dp[i][j]=1;
	    					if(i+1<n && j-1>=0 && dp[i+1][j-1]==1) dp[i][j]=1;
	    				}
	    				if((i==l||j==r) && dp[i][j]==1 && j-i+1>cnt) {
	    					cnt=j-i+1;
	    					tmp=new StringBuilder();
	    					for(int pos=i;pos<=j;pos++) tmp.append(a[pos]);
	    				}
	    			}
	    		}
	    		out.println(left+""+tmp+""+right);
	    	}
	        out.flush();
	        out.close();
	    }
	}