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;
long long check(long long a) {
  long long f = 0;
  for (long long i = 2; i * i <= a; i++)
    if (a % i == 0) {
      f = 1;
      break;
    }
  return f;
}
int32_t main() {
  long long n;
  cin >> n;
  for (long long i = 2;; i++) {
    long long a = i;
    long long b = i + n;
    if (check(a) && check(b)) return cout << b << " " << a << '\n', 0;
  }
}