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;
const int mod = 1e9 + 7;
const double pi = 3.14159265358979323846264;
void solve(int n) {
  int b, p, a = 0, s;
  scanf("%d%d", &b, &p);
  p *= n;
  while (n > 1) {
    s = 1;
    while (2 * s <= n) s *= 2;
    a += s / 2 * (2 * b + 1);
    n -= s / 2;
  }
  printf("%d %d", a, p);
}
int main() {
  int n, m, x, y, i, j, q = 0;
  while (scanf("%d", &n) != EOF) {
    solve(n);
    printf("\n");
  }
  return 0;
}
