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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.LinkedList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author KharYusuf
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        D1PrefixSuffixPalindromeEasyVersion solver = new D1PrefixSuffixPalindromeEasyVersion();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class D1PrefixSuffixPalindromeEasyVersion {
        public void solve(int testNumber, FastReader s, PrintWriter w) {
            char[] c = s.next().toCharArray();
            int n = c.length;
            int l = 0, r = n - 1;
            LinkedList<Character> fr = new LinkedList<>(), rev = new LinkedList<>();
            while (l <= r && c[l] == c[r]) {
                if (l != r) {
                    fr.add(c[l]);
                    rev.addFirst(c[r]);
                } else fr.add(c[l]);
                l++;
                r--;
            }
            int bestl = -1, bestr = -2, mx = 0;
            if (l <= r) {
                int len = r - l + 1;
                char[] pos1 = new char[len << 1 | 1], pos2 = new char[len << 1 | 1];
                for (int i = l, j = r; i <= r; i++, j--) {
                    pos1[i - l] = c[i];
                    pos2[i - l] = c[j];
                }
                pos1[len] = '*';
                pos2[len] = '*';
                for (int i = l, j = r; i <= r; i++, j--) {
                    pos1[len + i - l + 1] = c[j];
                    pos2[len + i - l + 1] = c[i];
                }
                //w.println(new String(pos1)+" "+new String(pos2));
                len <<= 1;
                len++;
                int[] lps1 = new int[len], lps2 = new int[len];
                computeLPSArray(pos1, len, lps1);
                computeLPSArray(pos2, len, lps2);
                if (lps1[len - 1] > mx) {
                    bestl = l;
                    bestr = l + lps1[len - 1] - 1;
                    mx = lps1[len - 1];
                }
                if (lps2[len - 1] > mx) {
                    bestl = r - lps2[len - 1] + 1;
                    bestr = r;
                }
            }
            for (int i = bestl; i <= bestr; i++) fr.add(c[i]);
            for (char cc : fr) w.print(cc);
            for (char cc : rev) w.print(cc);
            w.println();
        }

        void computeLPSArray(char[] pat, int M, int[] lps) {
            // length of the previous longest prefix suffix
            int len = 0;
            int i = 1;
            lps[0] = 0; // lps[0] is always 0

            // the loop calculates lps[i] for i = 1 to M-1
            while (i < M) {
                if (pat[i] == pat[len]) {
                    len++;
                    lps[i] = len;
                    i++;
                } else // (pat[i] != pat[len])
                {
                    // This is tricky. Consider the example.
                    // AAACAAAA and i = 7. The idea is similar
                    // to search step.
                    if (len != 0) {
                        len = lps[len - 1];

                        // Also, note that we do not increment
                        // i here
                    } else // if (len == 0)
                    {
                        lps[i] = len;
                        i++;
                    }
                }
            }
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private FastReader.SpaceCharFilter filter;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {

            if (numChars == -1)
                throw new InputMismatchException();

            if (curChar >= numChars) {

                curChar = 0;

                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }

                if (numChars <= 0)
                    return -1;
            }

            return buf[curChar++];
        }

        public String next() {

            int c = read();

            while (isSpaceChar(c))
                c = read();

            StringBuilder res = new StringBuilder();

            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {

            if (filter != null)
                return filter.isSpaceChar(c);

            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}