Problem: Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.

Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.

Now Pari asks you to write a program that gets a huge integer n from the input and tells what is the n-th even-length positive palindrome number?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10100 000).

Output

Print the n-th even-length palindrome number.

Examples

Input

1


Output

11


Input

10


Output

1001

Note

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

// author @mdazmat9
public class CodeForces_CA {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        String s=sc.next();
        StringBuilder t = new StringBuilder();
        t.append(s);
        t=t.reverse();
        System.out.print(s+t);




    }
    static int binarySearch(int arr[], int l, int r, int x)
    {
        if (r>=l)
        {
            int mid = l + (r - l)/2;

            // If the element is present at the
            // middle itself
            if (arr[mid] == x)
                return mid;

            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (arr[mid] > x)
                return binarySearch(arr, l, mid-1, x);

            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid+1, r, x);
        }

        // We reach here when element is not present
        //  in array
        return -1;
    }
    static long gcd(long a , long b)
    {
        if(b == 0)
            return a;
        return gcd(b , a % b);
    }

}
class Scanner {
    public BufferedReader reader;
    public StringTokenizer st;

    public Scanner(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        st = null;
    }

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

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

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

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

class OutputWriter {
    BufferedWriter writer;

    public OutputWriter(OutputStream stream) {
        writer = new BufferedWriter(new OutputStreamWriter(stream));
    }

    public void print(int i) throws IOException {
        writer.write(i);
    }

    public void print(String s) throws IOException {
        writer.write(s);
    }

    public void print(char[] c) throws IOException {
        writer.write(c);
    }

    public void close() throws IOException {
        writer.close();
    }

}