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:
# https://codeforces.com/contest/1269/problem/A
# Codeforces Round - 609 (Div. 2)
from math import sqrt
from math import ceil


def is_composite(num):
    for d in range(ceil(sqrt(num)), 1, -1):
        if num % d == 0:
            return True

    return False


n = int(input())
maximum = 10 ** 9
a_start = 5 if n <= 5 else n + 1
for a in range(a_start, maximum + 2):
    b_start = a - n
    if b_start <= 3:
        continue

    if is_composite(a):
        for b in range(b_start, a):
            if a - b != n:
                continue
            if is_composite(b):
                print(f'{a} {b}')
                exit()