#include <bits/stdc++.h>
using namespace std;
int main() {
  int p, k, s, x, y, i;
  int n, m, a, b, div;
  int costo1, costo2, costo3, menor;
  cin >> n >> m >> a >> b;
  costo1 = a * n;
  div = n / m;
  costo2 = div * b;
  div = n % m;
  costo2 = costo2 + (div * a);
  if (costo1 < costo2)
    menor = costo1;
  else
    menor = costo2;
  for (i = 1; i < 1000; i++) {
    if (i * m <= n) {
      costo3 = i * m;
      div = n - costo3;
      costo3 *= b;
      costo3 = costo3 + (div * a);
      if (costo3 < menor) menor = costo3;
    }
  }
  div = n / m;
  if (div * m < n) div++;
  costo3 = div * b;
  if (costo3 < menor) menor = costo3;
  cout << menor << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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>
using namespace std;
int main() {
  int p, k, s, x, y, i;
  int n, m, a, b, div;
  int costo1, costo2, costo3, menor;
  cin >> n >> m >> a >> b;
  costo1 = a * n;
  div = n / m;
  costo2 = div * b;
  div = n % m;
  costo2 = costo2 + (div * a);
  if (costo1 < costo2)
    menor = costo1;
  else
    menor = costo2;
  for (i = 1; i < 1000; i++) {
    if (i * m <= n) {
      costo3 = i * m;
      div = n - costo3;
      costo3 *= b;
      costo3 = costo3 + (div * a);
      if (costo3 < menor) menor = costo3;
    }
  }
  div = n / m;
  if (div * m < n) div++;
  costo3 = div * b;
  if (costo3 < menor) menor = costo3;
  cout << menor << endl;
  return 0;
}

```