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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
int main() {
  long long int n;
  cin >> n;
  long long int a;
  long long int b;
  long long int limit = (int)pow(10, 9);
  bool res_i = true;
  bool res_j = true;
  for (long long int i = 4; i < limit; ++i) {
    res_i = false;
    res_j = false;
    double sqrt_j = sqrt(n + i);
    double sqrt_i = sqrt(i);
    for (long long int k = 3; k <= sqrt_i; k += 2) {
      if (i % k == 0) {
        res_i = true;
        break;
      }
    }
    for (long long int l = 3; l <= sqrt_j; l += 2) {
      if ((n + i) % l == 0) {
        res_j = true;
        break;
      }
    }
    if (res_i == true && res_j == true) {
      cout << n + i << " " << i;
      break;
    } else {
      continue;
    }
  }
}