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 simp(long long ch) {
  for (int i = 2; i <= round(sqrt(ch)); i++) {
    if (ch % i == 0) {
      return true;
    }
  }
  return false;
}
void solve() {
  long long n;
  cin >> n;
  vector<int> vec;
  vec.push_back(4);
  vec.push_back(6);
  vec.push_back(9);
  vec.push_back(10);
  vec.push_back(12);
  vec.push_back(14);
  vec.push_back(15);
  vec.push_back(16);
  vec.push_back(18);
  vec.push_back(20);
  for (int i = 0; i < vec.size(); i++) {
    if (simp(n + vec[i])) {
      cout << n + vec[i] << " " << vec[i] << endl;
      return;
    }
  }
}
int main() {
  int q;
  solve();
}