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.io.BufferedWriter;
import java.io.File;
import java.io.PrintWriter;
import java.util.Scanner;

public class Main {

  // 1   - 9    -> 45
  // 10  - 99   -> pre + pre * 9 + pre * 10 = 900
  // 100 - 1000 -> 900 + (900 * 9) + (45 * 100) -> pre * 10 + 45 * 100
  
  private static BigInteger TEN = BigInteger.valueOf(10);
  private static BigInteger HUNDRED = BigInteger.valueOf(100);
  private static BigInteger FORTYFIVE = BigInteger.valueOf(45);
  
  private void solve() throws Exception {
    Scanner scanner = new Scanner(System.in);
    BigInteger k = scanner.nextBigInteger();
    
    BigInteger x = TEN.pow(100).subtract(BigInteger.ONE);
    BigInteger t = k.subtract(TEN.pow(99).multiply(FORTYFIVE.multiply(HUNDRED)).mod(k));
    System.out.println(t.toString() + " " + t.add(x).toString());
  }
  
  public static void main(String... args) {
  
    try {
      new Main().solve();
    } catch (Exception e) {
      e.printStackTrace();
    }
  
  }
}
