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() {
  long long int i, j, k, m, n, x, y;
  cin >> n;
  if (n == 1) {
    x = 9;
    y = 8;
  }
  if (n == 2) {
    x = 6;
    y = 4;
  }
  if (n == 3) {
    x = 9;
    y = 6;
  }
  if ((n != 2) && (n != 1) && (n != 3)) {
    if (n % 2 == 0) {
      x = 2 * n;
      y = n;
    } else {
      x = n + 9;
      y = 9;
    }
  }
  cout << x << " " << y << "\n";
}