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 os
import io
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#Fast input
"""
def flip(s,k):
    for i in range(len(s)-k+1):
        s[i:i+k] = s[i:i+k][::-1]
        #print(s[i:i+k], s[i:i+k][::-1])
    return s
t = int(input())
for i in range(t):
    n = int(input())
    s = list(input())
    a = ["z"]*n
    ans = None
    for k in range(1, n+1):
        c = flip(s[:], k)
        #print(c, k)
        if c < a:
            a = c
            ans = k
    print("".join(a))
    print(ans)"""

n = int(input())
if n%2==0:
    print(n+4, 4)
else:
    for i in range(1,10000,2):
        k = i**2
        if k > n and k-n > 2:
            print(k, k-n)
            break