Little X has met the following problem recently. 

Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>

Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code: 
    
    
      
        ans = solve(l, r) % a;  
        if (ans <= 0)  
          ans += a;  
      
    

This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.

Input

The first line contains a single integer a (1 ≤ a ≤ 1018).

Output

Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.

Examples

Input

46


Output

1 10


Input

126444381000032


Output

2333333 2333333333333

import java.math.BigInteger;
import java.util.Scanner;

/**
 *
 * @author RobertAlonso
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        BigInteger a = new BigInteger(sc.next());
        BigInteger g = BigInteger.valueOf(10);
        BigInteger p = BigInteger.valueOf(10);
        p = p.pow(18);
        g = g.pow(17); g = g.multiply(BigInteger.valueOf(45*18));
        
        /*BigInteger k = g.add(BigInteger.ONE);
        k = k.subtract(a); k = k.divide(a);
        if(k.mod(a)!=BigInteger.ZERO) k.add(BigInteger.ONE);
        k.multiply(BigInteger.valueOf(-1));        
        */        
        BigInteger k = ((a.subtract(BigInteger.ONE).subtract(g))).divide(a);
        if(k.mod(a)!=BigInteger.ZERO) k=k.subtract(BigInteger.ONE);
        
        BigInteger x = g.add(a.multiply(k));
        
        while((a.subtract(x)).compareTo(BigInteger.ONE) == -1){
            x = x.subtract(a);
        }
        
        /*System.out.println(a.subtract(BigInteger.ONE));
        System.out.println(g);
        System.out.println(g.add(a.multiply(k)));*/
        
        System.out.println(a.subtract(x) + " " + 
                (a.subtract(x)).add(p).subtract(BigInteger.ONE));
    }
    
}
