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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
bool check(int k) {
  int p = (int)ceil(sqrt(k));
  for (int i = 2; i <= p; ++i)
    if (k % i == 0) return 1;
  return 0;
}
int main() {
  cin >> N;
  int x = 200, y = x + N;
  for (;; ++x, ++y)
    if (check(x) && check(y)) return printf("%d %d\n", y, x), 0;
  return 0;
}