A tennis tournament with n participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.

The tournament takes place in the following way (below, m is the number of the participants of the current round):

  * let k be the maximal power of the number 2 such that k ≤ m, 
  * k participants compete in the current round and a half of them passes to the next round, the other m - k participants pass to the next round directly, 
  * when only one participant remains, the tournament finishes. 



Each match requires b bottles of water for each participant and one bottle for the judge. Besides p towels are given to each participant for the whole tournament.

Find the number of bottles and towels needed for the tournament.

Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).

Input

The only line contains three integers n, b, p (1 ≤ n, b, p ≤ 500) — the number of participants and the parameters described in the problem statement.

Output

Print two integers x and y — the number of bottles and towels need for the tournament.

Examples

Input

5 2 3


Output

20 15


Input

8 2 4


Output

35 32

Note

In the first example will be three rounds:

  1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 
  2. in the second round will be only one match, so we need another 5 bottles of water, 
  3. in the third round will also be only one match, so we need another 5 bottles of water. 



So in total we need 20 bottles of water.

In the second example no participant will move on to some round directly.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T abs(T a) {
  return ((a < 0) ? -a : a);
}
template <typename T>
inline T sqr(T a) {
  return a * a;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
  return out << "(" << p.first << ", " << p.second << ")";
}
bool solve(int);
int main() {
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(3);
  int test = 0;
  while (solve(test)) {
    ++test;
  }
  return 0;
}
int calc(int n, int b) {
  if (n == 1) return 0;
  int s = 1;
  while (s * 2 <= n) s *= 2;
  return s * b + (s / 2) + calc(s / 2 + n - s, b);
}
bool solve(int) {
  int n, b, p;
  if (scanf("%d%d%d", &n, &b, &p) != 3) return false;
  printf("%d %d\n", calc(n, b), n * p);
  return true;
}
