#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, a, b, sum1, sum2, temp1, temp2;
  while (scanf("%d %d %d %d", &n, &m, &a, &b) != EOF) {
    sum1 = n * a;
    if (n % m == 0) {
      sum2 = n / m * b;
    } else if (n >= m) {
      temp1 = (n / m + 1) * b;
      temp2 = n / m * b + n % m * a;
      sum2 = (temp1 < temp2 ? temp1 : temp2);
    } else if (n < m) {
      sum2 = b;
    }
    if (sum1 >= sum2) {
      printf("%d\n", sum2);
    } else {
      printf("%d\n", sum1);
    }
  }
  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>
using namespace std;
int main() {
  int n, m, a, b, sum1, sum2, temp1, temp2;
  while (scanf("%d %d %d %d", &n, &m, &a, &b) != EOF) {
    sum1 = n * a;
    if (n % m == 0) {
      sum2 = n / m * b;
    } else if (n >= m) {
      temp1 = (n / m + 1) * b;
      temp2 = n / m * b + n % m * a;
      sum2 = (temp1 < temp2 ? temp1 : temp2);
    } else if (n < m) {
      sum2 = b;
    }
    if (sum1 >= sum2) {
      printf("%d\n", sum2);
    } else {
      printf("%d\n", sum1);
    }
  }
  return 0;
}

```