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;
inline void read(int &x) {
  x = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c - '0');
}
inline void write(int x) {
  register char buffor[35];
  register int i = 0;
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
int main() {
  int t;
  {
    int n, b, p;
    read(n), read(b), read(p);
    int k, ans, fans = 0;
    int g = n;
    while (n > 1) {
      for (int i = 0; i < 32; i++) {
        if (n & (1 << (31 - i))) {
          ans = (1 << (31 - i));
          break;
        }
      }
      k = ans;
      fans += k * b + k / 2;
      n = (k / 2) + (n - k);
    }
    printf("%d %d\n", fans, p * g);
  }
  return 0;
}
