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.util.*;
import java.io.*;
import java.math.*;

// 似乎在萝莉控的ppt里看过题解…
// 先选1~10^18，然后接下来每次把左端和右端同时加1就相当于把答案加1(加了1000..000**, 删了**)

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		BigInteger a = in.nextBigInteger();
		BigInteger sum =
			BigInteger.valueOf(18).multiply(
				BigInteger.TEN.pow(18)
					.divide(BigInteger.TEN)
					.multiply(BigInteger.valueOf((0 + 9) * 10 / 2))
			).add(BigInteger.ONE).mod(a);
		BigInteger ans = a.subtract(sum).mod(a);
		System.out.print(BigInteger.ONE.add(ans));
		System.out.print(" ");
		System.out.print(BigInteger.TEN.pow(18).add(ans));
		System.out.println();
	}
}