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;
const int MAXN = 1;
int x;
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
int main() {
  x = read();
  printf("%d %d\n", x * 9, x * 8);
}