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;
bool p(long long n) {
  int d = 0;
  for (long long i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      d++;
      if (n / i != i) d++;
    }
  }
  return d == 2;
}
int main() {
  long long n, a, b;
  cin >> n;
  b = n;
  a = b + n;
  while (1) {
    if (!p(a) && !p(b)) break;
    a++;
    b++;
  }
  cout << a << ' ' << b << endl;
  return 0;
}