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 j = 8;
  int ans = 0, num = 0;
  while (ans == 0) {
    int check1 = 0, check2 = 0;
    num = j + n;
    for (int i = 1; i <= j; i++) {
      if (j % i == 0) check1++;
    }
    for (int i = 1; i <= num; i++) {
      if (num % i == 0) check2++;
    }
    if (check1 > 2 && check2 > 2) {
      ans = num;
      break;
    }
    j++;
  }
  cout << num << " " << j << endl;
  return 0;
}