#include <bits/stdc++.h>
int main() {
  int t, i, n, m, a, b;
  scanf("%d%d%d%d", &n, &m, &a, &b);
  if (a * m <= b) {
    printf("%d\n", n * a);
  } else if ((b * (n / m) + a * (n % m)) > (b * ((n / m) + 1))) {
    printf("%d\n", b * (n / m) + b);
  } else
    printf("%d\n", b * (n / m) + a * (n % m));
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

Ann has recently started commuting by subway. We know that a one ride subway ticket costs a rubles. Besides, Ann found out that she can buy a special ticket for m rides (she can buy it several times). It costs b rubles. Ann did the math; she will need to use subway n times. Help Ann, tell her what is the minimum sum of money she will have to spend to make n rides?

Input

The single line contains four space-separated integers n, m, a, b (1 ≤ n, m, a, b ≤ 1000) — the number of rides Ann has planned, the number of rides covered by the m ride ticket, the price of a one ride ticket and the price of an m ride ticket. 

Output

Print a single integer — the minimum sum in rubles that Ann will need to spend.

Examples

Input

6 2 1 2


Output

6


Input

5 2 2 3


Output

8

Note

In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three m ride tickets.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int t, i, n, m, a, b;
  scanf("%d%d%d%d", &n, &m, &a, &b);
  if (a * m <= b) {
    printf("%d\n", n * a);
  } else if ((b * (n / m) + a * (n % m)) > (b * ((n / m) + 1))) {
    printf("%d\n", b * (n / m) + b);
  } else
    printf("%d\n", b * (n / m) + a * (n % m));
  return 0;
}

```