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 lalala(int k) {
  int b, c, d, i;
  for (i = k / 2; i > 1; i--) {
    if (k % i == 0) break;
  }
  if (i == 1)
    return 0;
  else
    return 1;
}
int main(void) {
  int a, b, x, m, n;
  scanf("%d", &n);
  int i, j;
  for (i = 4; i <= 109; i++) {
    a = i;
    b = i + n;
    if (lalala(a) == 1 && lalala(b) == 1) break;
  }
  printf("%d %d", b, a);
  return 0;
}