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;
bool is_prime(int n) {
  int i;
  bool isPrime = true;
  for (i = 2; i <= n / 2; ++i) {
    if (n % i == 0) {
      isPrime = false;
      break;
    }
  }
  return isPrime;
}
int main() {
  int number;
  cin >> number;
  for (int b = 2;; b++) {
    int c = 0;
    if (!is_prime(b)) {
      int a = b + number;
      if (!is_prime(a)) {
        cout << a << " " << b << "\n";
        c = 1;
      }
    }
    if (c == 1) break;
  }
}