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>
int f1(long long int n) {
  long long int a, count = 0;
  for (a = 2; a < n; a++) {
    if (n % a == 0) count++;
  }
  return (count >= 1);
}
int main() {
  long long int n;
  scanf("%lld", &n);
  long long int a, b = 4;
  while (1) {
    a = n + b;
    if (f1(a) && f1(b)) {
      printf("%lld %lld\n", a, b);
      break;
    }
    b++;
  }
}