Problem: You are given a string A. Find a string B, where B is a palindrome and A is a subsequence of B.

A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".

A palindrome is a string that reads the same forward or backward.

The length of string B should be at most 104. It is guaranteed that there always exists such string.

You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104.

Input

First line contains a string A (1 ≤ |A| ≤ 103) consisting of lowercase Latin letters, where |A| is a length of A.

Output

Output single line containing B consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104. If there are many possible B, print any of them.

Examples

Input

aba


Output

aba

Input

ab


Output

aabaa

Note

In the first example, "aba" is a subsequence of "aba" which is a palindrome.

In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.

Solution (in JAVA):

Solution:
import java.io.ByteArrayInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;
import java.util.Queue;
import java.util.logging.Level;
import java.util.logging.Logger;

/*
 *
 * @author ssaxena36 :D
 *
 */
public class NewClass {
    class Node{
        int data;
        Node next;
        Node(int data){
            this.data=  data;
            next = null;
        }
    }
    void solve() {
        String string = inps();
        out.print(string);
        for(int i = string.length() - 1; i >= 0; i--){
            out.print(string.charAt(i));
        }
    }

    public boolean isPrime(long n) {
        return BigInteger.valueOf(n).isProbablePrime(5);
    }

    public boolean[] SieveOfEratosthenes(int n) {
        boolean prime[] = new boolean[n + 1];
        Arrays.fill(prime, true);
        for (int p = 2; p * p <= n; p++) {
            if (prime[p] == true) {
                for (int i = p * 2; i <= n; i += p) {
                    prime[i] = false;
                }
            }
        }
        for (int i = 2; i < n; i++) {
            if (prime[i]) {
                //do some stuff
            }
        }
        return prime;
    }
    InputStream obj;
    PrintWriter out;
    String check = "";
    int len;

    public static void main(String[] args) throws IOException {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    new NewClass().main1();
                } catch (IOException ex) {
                    Logger.getLogger(NewClass.class.getName()).log(Level.SEVERE, null, ex);
                }
            }
        }, "child-thread", 1 << 26).start();

    }

    void main1() throws IOException {
        out = new PrintWriter(System.out);
        obj = check.isEmpty() ? System.in : new ByteArrayInputStream(check.getBytes());
        //   obj=check.isEmpty() ? new FileInputStream("/home/ssaxena36/Desktop/file.in") : new ByteArrayInputStream(check.getBytes());
        solve();
        out.flush();
        out.close();
    }
    byte inbuffer[] = new byte[1024];
    int lenbuffer = 0, ptrbuffer = 0;

    int readByte() {
        if (lenbuffer == -1) {
            throw new InputMismatchException();
        }
        if (ptrbuffer >= lenbuffer) {
            ptrbuffer = 0;
            try {
                lenbuffer = obj.read(inbuffer);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }
        if (lenbuffer <= 0) {
            return -1;
        }
        return inbuffer[ptrbuffer++];
    }

    boolean isSpaceChar(int c) {
        return (!(c >= 33 && c <= 126));
    }

    int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b));
        return b;
    }

    String inps() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    int inpi() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    long inpl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    float inpf() {
        return Float.parseFloat(inps());
    }

    double inpd() {
        return Double.parseDouble(inps());
    }

    char inpc() {
        return (char) skip();
    }

    int[] inpia(int n) {
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = inpi();
        }
        return a;
    }

    long[] inpla(int n) {
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = inpl();
        }
        return a;
    }

    String[] inpsa(int n) {
        String a[] = new String[n];
        for (int i = 0; i < n; i++) {
            a[i] = inps();
        }
        return a;
    }

    double[][] inpdm(int n, int m) {
        double a[][] = new double[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = inpd();
            }
        }
        return a;
    }

    int[][] inpim(int n, int m) {
        int a[][] = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = inpi();
            }
        }
        return a;
    }
}