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;
struct piii {
  long long first, second, th;
  bool operator<(const piii &b) {
    if (first != b.first) return first < b.first;
    if (second != b.second) return second < b.second;
    return th < b.th;
  }
};
int main() {
  cout << setprecision(2) << fixed;
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  cout << 9 * n << ' ' << 8 * n;
  return 0;
}