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 f(int n) {
  int i, flag = 0;
  if (n == 1)
    flag = 0;
  else if (n == 2)
    flag = 0;
  else {
    int j = sqrt(n + 0.5);
    for (i = 2; i < j; i++) {
      if (n % i == 0) {
        flag = 1;
        break;
      }
    }
  }
  return flag;
}
int main() {
  int n;
  long long i;
  long long j = pow(10, 9);
  while (~scanf("%d", &n)) {
    for (i = 1; i <= j; i++) {
      if (f(i) == 1 && f(i + n) == 1) {
        printf("%lld %lld\n", i + n, i);
        break;
      }
    }
  }
}