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;
using ll = long long;
using ull = unsigned long long;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
using ld = long double;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T sqr(T x) {
  return x * x;
}
const ld pi = 4 * atan(1.);
const int inf = 1e9 + 7;
const int N = 3e5 + 17;
ll n, m, k;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cout << 9 * n << " " << 8 * n << '\n';
}