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;
map<int, pair<bool, bool> > prime;
long long n, a, b;
bool ThisisCheck(int a) {
  if (a < 2) return false;
  if (a == 2 || a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  long long k = 5;
  while (k <= sqrt(a)) {
    if (a % k == 0 || a % (k + 2) == 0) return false;
    k += 6;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  b = 4;
  while (true) {
    if (prime[b].second == 0) {
      if (ThisisCheck(b)) {
        b++;
        prime[b] = pair<bool, bool>(1, 1);
        continue;
      }
      prime[b] = pair<bool, bool>(0, 1);
    }
    if (prime[b].first == 1) {
      b++;
      continue;
    }
    prime[b + n].first = ThisisCheck(b + n);
    prime[b + n].second = 1;
    if (!prime[b + n].first) {
      cout << b + n << " " << b;
      return 0;
    }
    b++;
  }
  cout << b + n << " " << b;
  return 0;
}