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;
vector<long long int> divisor;
void getDivisors(long long int n) {
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        divisor.push_back(i);
      else {
        divisor.push_back(i);
        divisor.push_back(n / i);
      }
    }
  }
}
long long int power(long long int a, long long int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  long long int temp = power(a, n / 2);
  if (n % 2 == 0) {
    return ((temp % 1000000007) * (temp % 1000000007)) % 1000000007;
  } else {
    return ((((temp % 1000000007) * (temp % 1000000007)) % 1000000007) * a) %
           1000000007;
  }
}
long long int prime(long long int x) {
  for (int i = 2; i <= sqrt(x); i = i + 1)
    if (x % i == 0) return 1;
  return 0;
}
int main() {
  {
    long long int n;
    cin >> n;
    long long int x = 4;
    while (true) {
      if (prime(x) and prime(x + n)) {
        cout << x + n << " " << x << endl;
        return 0;
      }
      x++;
    }
  }
}