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>
long long n, prime[10000001], tot1;
bool p[100000001];
int main() {
  p[0] = p[1] = 1;
  for (long long i = 2; i < 100000001; i++) {
    if (!p[i]) prime[tot1++] = i;
    for (int j = 0; j < tot1 && i * prime[j] < 100000001; j++) {
      p[i * prime[j]] = 1;
      if (!(i % prime[j])) break;
    }
  }
  scanf("%lld", &n);
  for (long long i = 4; i < 100000001; i++) {
    if (p[i + n] && p[i]) {
      printf("%lld %lld", i + n, i);
      return 0;
    }
  }
  return 0;
}