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 n;
using namespace std;
const int times = 20;
int number = 0;
map<long long, int> m;
long long Random(long long n) { return ((double)rand() / RAND_MAX * n + 0.5); }
long long q_mul(long long a, long long b, long long mod) {
  long long ans = 0;
  while (b) {
    if (b & 1) {
      b--;
      ans = (ans + a) % mod;
    }
    b /= 2;
    a = (a + a) % mod;
  }
  return ans;
}
long long q_pow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = q_mul(ans, a, mod);
    }
    b /= 2;
    a = q_mul(a, a, mod);
  }
  return ans;
}
bool witness(long long a, long long n) {
  long long tem = n - 1;
  int j = 0;
  while (tem % 2 == 0) {
    tem /= 2;
    j++;
  }
  long long x = q_pow(a, tem, n);
  if (x == 1 || x == n - 1) return true;
  while (j--) {
    x = q_mul(x, x, n);
    if (x == n - 1) return true;
  }
  return false;
}
bool miller_rabin(long long n) {
  if (n == 2) return true;
  if (n < 2 || n % 2 == 0) return false;
  for (register int i = 1; i <= times; i++) {
    long long a = Random(n - 2) + 1;
    if (!witness(a, n)) return false;
  }
  return true;
}
int main() {
  cin >> n;
  for (int i = 1000000000; i >= n + 1; i--) {
    if (!miller_rabin(i) && !miller_rabin(i - n)) {
      cout << i << " " << i - n;
      return 0;
    }
  }
  return 0;
}