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:
#code
def iscomposite(a):
    count=0
    for i in range(2,a):
        if(count==1):
            break
        if(a%i==0):
            count+=1 
    if(count>0):
        return True
    else:
        return False
n=int(input())
#print(iscomposite(n))
if(n==1):
    print(9,8)
elif(n==2):
    print(6,4)
elif(iscomposite(n)):
    print(2*n,n)
else:
    p=0
    p=n+1
    while(iscomposite(p)==False or iscomposite(p+n)==False):
        p+=1 
    print(p+n,p)