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 m = 1000000007;
long long sieve[1000001] = {0}, i, j;
long long power(long long p, long long k) {
  long long a = 1;
  while (k) {
    if (k & 1) a = (a * p) % m;
    p = (p * p) % m;
    k >>= 1;
  }
  return a;
}
long long prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 1;
  return 0;
}
using namespace std;
int main() {
  long long tc = 1;
  long long k, n;
  cin >> n;
  k = n;
  for (i = 2; i <= 10000000; i++)
    if (prime(i) && prime(i + k)) {
      cout << i + n << " " << i;
      break;
    }
}