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.*;
import java.util.*;

public class D implements Runnable {
	FastReader sc;
	PrintWriter out;
	long mod = (long)1e9 + 7;
	int maxA = (int)1e7+5;
	
	boolean isPal(String s,int l,int r)
	{	
		int j = r;
		for(int i = l; i <= r; i++)
		{
			if(s.charAt(i) != s.charAt(j))
			{
				return false;
			}
			j--;
		}
		
		return true;
	}
	
	String reverse(String s)
	{
		StringBuffer sb = new StringBuffer(s);
		sb.reverse();
		return sb.toString();
	}
	
	int[] prefixFunction(String s)
	{
		int n = s.length();
		int pi[] = new int[n];
		
		pi[0] = 0;
		
		for(int i = 1; i < n; i++)
		{
			int j = pi[i-1];
			
			while(j > 0 && s.charAt(i) != s.charAt(j))
			{
				j = pi[j-1];
			}
			
			if(s.charAt(i) == s.charAt(j))
			{
				j++;
			}
			
			pi[i] = j;
		}
		return pi;
	}
	
	void solve() 
	{		
		
		int tc = sc.ni();
		outer: while(tc-- > 0)
		{
			String s = sc.nln();
			
			int l = 0;
			int r = s.length()-1;
			
			while(l < r && s.charAt(l) == s.charAt(r))
			{
				l++;
				r--;
			}
			
			if(l == r)
			{
				out.println(s);
				continue outer;
			}
			
			String s1 = s.substring(l,r+1);
			String s2 = reverse(s1);
			
			int n = s1.length();
			
			int[] pf1 = prefixFunction(s1+"#"+s2);
			int[] pf2 = prefixFunction(s2+"#"+s1);

			
			int la = 0;
			int ra = 0;
			
			int n1 = pf1.length;
			
			la = pf1[n1-1];
			ra = pf2[n1-1];

			
			
			
			StringBuffer sb = new StringBuffer();
			if(l != 0)
			{
				sb = new StringBuffer(s.substring(0,l));
			}
			
			if(la > ra)
			{
				sb.append(s.substring(l,l+la));
			}
			else
			{
				sb.append(s.substring(r-ra+1,r+1));
			}

			if(l != 0)
			{
				sb.append(s.substring(r+1,s.length()));
			}
			out.println(sb.toString());
		}
	}
	
	
	
	class Pair implements Comparable<Pair>
	{
		
		int x;
		int y;
		
		Pair(int x,int y)
		{
			this.x = x;
			this.y = y;
		}
		
		public int compareTo(Pair o) 
		{
			
			if(x != o.x)
			{
				if(x > o.x) return 1;  //increasing order
				else return -1;	
			}
			else
			{
				if(y > o.y) return 1;  //increasing order
				else return -1;	
			}
		}
	}

	public void run() 
	{
		out = new PrintWriter(System.out);
		sc = new FastReader();
		solve();
		out.flush();
	}

	public static void main(String[] args) 
	{
		new Thread(null, new D(), "Main", 1 << 28).start();
	}

	class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            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 ni() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nl() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nd() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nln() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
        

		int[] nia(int n) {
			int[] ar = new int[n];
			for (int i = 0; i < n; i++)
				ar[i] = ni();
			return ar;
		}
		
		long[] nla(int n) {
			long[] ar = new long[n];
			for (int i = 0; i < n; i++)
				ar[i] = nl();
			return ar;
		}

	    
		int[][] nim(int n, int m) {
			int[][] ar = new int[n][];
			for (int i = 0; i < n; i++) {
				ar[i] = nia(m);
			}
			return ar;
		}

		long[][] nlm(int n, int m) {
			long[][] ar = new long[n][];
			for (int i = 0; i < n; i++) {
				ar[i] = nla(m);
			}
			return ar;
		}
		
		String reverse(String s)
		{
			StringBuilder r = new StringBuilder(s);
			return r.reverse().toString();
		}
		
		void merge1(int arr[], int l, int m, int r)
	    {
	        int n1 = m - l + 1;
	        int n2 = r - m;
	        int L[] = new int [n1];
	        int R[] = new int [n2];
	        for (int i=0; i<n1; ++i)
	            L[i] = arr[l + i];
	        for (int j=0; j<n2; ++j)
	            R[j] = arr[m + 1+ j];
	        int i = 0, j = 0;
	        int k = l;
	        while (i < n1 && j < n2)
	        {
	            if (L[i]<=R[j])
	            {
	                arr[k] = L[i];
	                i++;
	            }
	            else
	            {
	                arr[k] = R[j];
	                j++;
	            }
	            k++;
	        }
	        while (i < n1)
	        {
	            arr[k] = L[i];
	            i++;
	            k++;
	        }
	        while (j < n2)
	        {
	            arr[k] = R[j];
	            j++;
	            k++;
	        }
	    }
	    void sort1(int arr[], int l, int r)
	    {
	        if (l < r)
	        {
	            int m = (l+r)/2;
	            sort1(arr, l, m);
	            sort1(arr , m+1, r);
	            merge1(arr, l, m, r);
	        }
	    }
    }
}