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;
int main(void) {
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n + 4 << " " << 4;
  } else {
    if (n == 1) {
      cout << 9 << " " << 8;
    } else if (n == 3) {
      cout << 12 << " " << 9;
    } else if (n == 5) {
      cout << 15 << " " << 10;
    } else if (n == 7) {
      cout << 16 << " " << 9;
    } else if (n == 9) {
      cout << 15 << " " << 6;
    } else {
      if (n % 10 < 5)
        cout << 2 * n + (5 - n % 5) << " " << n + (5 - n % 5);
      else
        cout << 2 * n - n % 5 << " " << n - n % 5;
    }
  }
  return 0;
}