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;
int fx4[] = {1, -1, 0, 0};
int fy4[] = {0, 0, 1, -1};
int Kfx[] = {-2, -2, 2, 2, 1, 1, -1, -1};
int Kfy[] = {1, -1, 1, -1, 2, -2, 2, -2};
int fx8[] = {1, 1, 1, 0, 0, -1, -1, -1};
int fy8[] = {0, 1, -1, 1, -1, 0, 1, -1};
long long int BM(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  long long int x = BM(a, b / 2, m);
  x = ((x % m) * (x % m)) % m;
  if (b % 2) x = ((x % m) * (a % m)) % m;
  return x;
}
bool p(int v) {
  int f = 0;
  for (int i = 2; i <= sqrt(v); i++) {
    if (v % i == 0) {
      f = 1;
      break;
    }
  }
  return f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 4;; i++) {
    if (p(i) == 1 and p(i + n) == 1) {
      cout << i + n << " " << i << endl;
      break;
    }
  }
}