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 modexpo(long long x, long long n, long long m) {
  if (n == 0) return 1;
  long long tmp = modexpo(x, n / 2, m);
  tmp = (tmp * tmp) % m;
  if (n & 1) tmp = (x * tmp) % m;
  return tmp;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  cout << 15 * n << " " << 14 * n;
}