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;
long long int flag[19999999];
void seive() {
  for (int i = 4; i <= 19999999; i += 2) {
    flag[i] = 1;
  }
  for (int i = 3; i <= sqrt(19999999); i += 2) {
    if (flag[i] == 0) {
      for (int j = i * i; j <= 19999999; j += i) flag[j] = 1;
    }
  }
}
int main() {
  int n;
  cin >> n;
  seive();
  for (int i = 1; i <= 19999999; i++) {
    long long int b = i;
    long long int a = n + b;
    if (flag[a] == 1 && flag[b] == 1) {
      cout << a << ' ' << b << endl;
      break;
    }
  }
}