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:
n = int(input())
counter = 0
sounter = 0
num1 = 8
num2 = num1 + n
while(True):
    for i in range(2, num1):
        if num1 % i == 0:
            counter += 1
    for j in range(2, num2):
        if num2 % j == 0:
            sounter += 1
    if counter != 0 and sounter != 0:
        print(num2, num1, sep=" ")
        exit()
    else:
        num1 += 1
        num2 += 1
        counter = 0
        sounter = 0