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.StringTokenizer;

public class E {

    static class Task {

        private StrHash strHash;
        private StrHash reverseStrHash;

        private boolean isPalindrome(char[] arr, int left, int right) {
            int len = arr.length;
            int count = right - left + 1;
            int mid = (left + right) >>> 1;
            long leftHash;
            long rightHash;
            if (count % 2 == 0) {
                leftHash = strHash.getSubStrHash(left, mid);
                rightHash = reverseStrHash.getSubStrHash(len - right - 1, len - mid - 2);
            } else {
                leftHash = strHash.getSubStrHash(left, mid - 1);
                rightHash = reverseStrHash.getSubStrHash(len - right - 1, len - mid - 2);
            }
            return leftHash == rightHash;
        }

        // 左边固定
        private int[] getMaxPalindromeLeftToRight(char[] arr, int from, int to) {
            int left = from;
            for (int right = to; right > left; right--) {
                if (isPalindrome(arr, left, right)) {
                    return new int[]{left, right};
                }
            }
            return new int[] {left, left};
        }

        // 右边固定
        private int[] getMaxPalindromeRightToLeft(char[] arr, int from, int to) {
            int right = to;
            for (int left = from; left < right; left++) {
                if (isPalindrome(arr, left, right)) {
                    return new int[]{left, right};
                }
            }
            return new int[]{to, to};
        }

        private void calcStrHash(String str) {
            String reverseStr = new StringBuilder(str).reverse().toString();
            strHash = new StrHash(str.toCharArray());
            reverseStrHash = new StrHash(reverseStr.toCharArray());
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int t = in.nextInt();
            while ((t--) > 0) {
                String str = in.next();
                char[] arr = str.toCharArray();

                // 两个方向都要判断
                int len = arr.length;
                int left = 0;
                int right = len - 1;

                while (left < right && arr[left] == arr[right]) {
                    left++;
                    right--;
                }

                if (left >= right) {
                    out.println(str);
                    continue;
                }

                // 从左边找最大的 和从右边找最大的回文串
                calcStrHash(str);
                int[] leftToRight = getMaxPalindromeLeftToRight(arr, left, right);
                int[] rightToLeft = getMaxPalindromeRightToLeft(arr, left, right);
                String ans = "";
                if (leftToRight[1] - leftToRight[0] > rightToLeft[1] - rightToLeft[0]) {
                    ans = str.substring(0, leftToRight[1] + 1);
                    if (right < len - 1) {
                        ans += str.substring(right + 1, len);
                    }
                } else {
                    if (left > 0) {
                        ans = str.substring(0, left);
                    }
                    ans = ans + str.substring(rightToLeft[0]);
                }

                out.println(ans);
            }
        }
    }

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task task = new Task();
        task.solve(1, in, out);
        out.close();
    }

    static class StrHash {

        private static final long P = 805306457;
        private static final long MOD = (int) (1e9+7);
        private long[] hash;
        private long[] power;
        private char[] arr;

        public StrHash(char[] arr) {
            this.arr = arr;
            int len = arr.length;
            hash = new long[len];
            power = new long[len];
            calcHashAndPower();
        }

        private void calcHashAndPower() {
            hash[0] = arr[0];
            power[0] = 1;
            for (int i = 1; i < arr.length; i++) {
                hash[i] = (hash[i-1] * P + arr[i]) % MOD;
                power[i] = (power[i-1] * P) % MOD;
            }
        }

        private long getSubStrHash(int l, int r) {
            if (l == 0) {
                return hash[r];
            }
            return ((hash[r] - power[r-l+1] * hash[l-1]) % MOD + MOD) % MOD;
        }

    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }

}