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() {
  int n;
  cin >> n;
  int i = 4;
  bool found = false;
  while (!found) {
    int a = i, b = i + n;
    bool checkA = false, checkB = false;
    for (int j = 2; j <= ceil(b / 2); j++) {
      if (b % j == 0) {
        checkB = true;
        break;
      }
    }
    for (int j = 2; j <= ceil(a / 2); j++) {
      if (a % j == 0) {
        checkA = true;
        break;
      }
    }
    if (checkA && checkB) {
      if (a > b)
        cout << a << ' ' << b;
      else
        cout << b << ' ' << a;
      found = true;
    }
    i++;
  }
}