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:
def prm(n):
    if n<=3:
        return True
    ans=True
    c=0
    for i in range(2,int((n)**(0.5))+1):
        if n%i==0:
            ans=False
            return ans
    return ans
n=int(input())
if n%2==0:
    print(4+n,4)
else:
    if abs(9-n)<=9:
        for i in range(9,10**9+1):
            if prm(i)==False:
                if prm(abs(n-i))==False:
                    if abs(n-i)>=4:
                        if abs(i-abs(n-i)==n):
                            print(max(i,abs(n-i)),min(i,abs(n-i)))
                            break
    else:
        for i in range(n+4,10**9+1):
            if prm(i)==False:
                if prm(abs(n-i))==False:
                    if abs(n-i)>=4:
                        if abs(i-abs(n-i))==n:
                            print(max(i,abs(n-i)),min(i,abs(n-i)))
                            break