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

/**
 * Created by user1 on 20.09.2014.
 */
public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        BigInteger N = in.nextBigInteger();
        BigInteger value = BigInteger.valueOf(45*51);
        for(int i = 0; i < 50; ++i){
            value = value.multiply(BigInteger.TEN);
        }
        //out.println(value);
        BigInteger mod = (N.multiply(BigInteger.valueOf(40)).subtract(value.mod(N))).mod(N).add(N);
        BigInteger right = mod.subtract(BigInteger.ONE);
        BigInteger left = mod;
        BigInteger poww = BigInteger.ONE;
        for(int i = 0; i < 51; ++i){
            poww = poww.multiply(BigInteger.TEN);
        }
        right = right.add(poww);
        out.print(left +" "+right);
        in.close();
        out.flush();
        out.close();

    }
}
