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 prime[100001];
vector<int> v;
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  v.push_back(2);
  for (int i = 3; i * i <= 100000; i += 2) {
    if (prime[i] == false) continue;
    v.push_back(i);
    for (int j = i * 2; j <= 100000; j += i) prime[j] = false;
  }
}
bool isprime(long long int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  sieve();
  int arr[n + 1];
  for (int i = 0; i < v.size(); i++) {
    int l = v[i] * 2;
    long long int r = n + l;
    if (!isprime(r)) {
      cout << r << " " << l;
      break;
    }
  }
}