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;
const long long N = 1e6 + 5;
long long pww(long long a, long long b) {
  if (b == 0) return 1;
  long long x = pww(a, b / 2);
  if (b % 2) return a * x * x;
  return x * x;
}
int n;
int main() {
  cin >> n;
  if (n == 1)
    cout << 9 << " " << 8 << '\n';
  else
    cout << 3 * n << " " << 2 * n << '\n';
  return 0;
}