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 modl(string num, int a) {
  int res = 0;
  for (int i = 0; i < num.length(); i++)
    res = (res * 10 + (int)num[i] - '0') % a;
  return res;
}
int powm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long m, n, i, j, k, p = 0, q = 0, r = 0, temp, flag = 0, x, y, z;
  int p1 = 0, q1 = 0, r1 = 0;
  string s, s2, s3;
  vector<char> v1, v2;
  cin >> n;
  if (n % 2 == 0)
    cout << n + 8 << " " << 8;
  else
    cout << n + 9 << " " << 9;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}