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 Composite(n):
    if n == 1:
        return False
    
    for i in range(2, int(math.ceil(n**0.5))+1):
        if n % i == 0:
            return True
    return False


if __name__ == "__main__":
    n =  int(input())
    b = 4
    while True:
        # print(n+b, b)
        if Composite(b) and Composite(n+b):
            print(n+b, b)
            break
        else:
            b += 1