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 f[1000001];
int prime[1000001];
void sieve() {
  prime[0] = prime[1] = 0;
  for (int i = 2; i <= 1000000; i++) prime[i] = 1;
  for (int i = 2; i * i <= 1000000; i++) {
    if (prime[i]) {
      for (long long int j = i * i; j <= 1000000; j += i) prime[j] = 0;
    }
  }
}
int power(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) {
      r = (r * 1ll * a) % 1000000007;
      b--;
    } else {
      a = (a * 1ll * a) % 1000000007;
      b >>= 1;
    }
  }
  return r;
}
void factorial() {
  f[0] = f[1] = 1;
  for (int i = 2; i <= 1000000; i++) f[i] = (f[i - 1] * 1ll * i) % 1000000007;
}
int nCr(int a, int b) {
  factorial();
  int r = f[a];
  r = (r * 1ll * power(f[b], 1000000007 - 2)) % 1000000007;
  r = (r * 1ll * power(f[a - b], 1000000007 - 2)) % 1000000007;
  return r;
}
vector<long long int> z_algo(string s) {
  long long int n = s.length(), l = 0, r = 0;
  vector<long long int> z(n);
  for (int i = 1; i <= n - 1; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (z[i] + i < n && s[z[i]] == s[i + z[i]]) z[i]++;
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  cout << 9 * n << " " << 8 * n << "\n";
  return 0;
}