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:
import math

def divisor(n):
    large_divisors = []
    for k in range(1, int(math.sqrt(n) + 1)):
        if(n % k == 0):
            yield k
            if(k*k != n):
                large_divisors.append(n//k)
    for d in reversed(large_divisors):
        yield d

n = int(input())
a = 100+n
b = 100

while(len(list(divisor(a))) == 2 or len(list(divisor(a)))==2):
    a -= 1
    b -= 1



print("{} {}".format(a, (b)))