Problem: Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:

  * the following numbers are composite: 1024, 4, 6, 9; 
  * the following numbers are not composite: 13, 1, 2, 3, 37. 



You are given a positive integer n. Find two composite integers a,b such that a-b=n.

It can be proven that solution always exists.

Input

The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.

Output

Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).

It can be proven, that solution always exists.

If there are several possible solutions, you can print any. 

Examples

Input


1


Output


9 8


Input


512


Output


4608 4096

Solution (in PYTHON3):

Solution:
#!/usr/bin/env python3
import sys
input = sys.stdin.readline

def prime_factorize(n, unique=False):
    b = 2
    fct = []
    while b * b <= n:
        while n % b == 0:
            n //= b
            fct.append(b)
        b = b + 1
    if n > 1:
        fct.append(n)
    if unique:
        return set(fct)
    else:
        return fct

n = int(input())
for i in range(4, 1000):
    if len(prime_factorize(i)) != 1:
        if len(prime_factorize(i+n)) != 1:
            print(i+n, i)
            exit()