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 mod = 1e9 + 7;
long long mode(long long n) { return ((n % mod) + mod) % mod; }
void solve() {
  long long n;
  cin >> n;
  cout << 9 * n << ' ' << 8 * n << endl;
}
void abc() {
  int i, j, k;
  int n, m;
  int x, y;
  cin >> n;
  long long cnt = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      x = i ^ j;
      if (x <= n && x > 0) {
        k = 0;
        if (i + j <= x)
          k++;
        else if (i + x <= j)
          k++;
        else if (x + j <= i)
          k++;
        if (k <= 0) cnt++;
      }
    }
  }
  cout << cnt / 6;
}
void bcd() {
  string s, ss;
  cin >> s >> ss;
  int sum = 0, n = s.size(), nn = ss.size();
  for (int i = 0; i <= n - nn; i++) {
    int a = 0;
    for (int j = i; j < i + nn; j++) {
      if (s[j] == ss[j - i]) {
        a++;
      }
    }
    sum = max(a, sum);
  }
  cout << ss.size() - sum << endl;
}
int main() {
  solve();
  return 0;
}