#include <bits/stdc++.h>
long long int min(long long int m1, long long int m2) {
  if (m1 < m2)
    return m1;
  else
    return m2;
}
int main() {
  long long int n, m, a, b, x, y, max = 1000000, i;
  scanf("%lld %lld %lld %lld", &n, &m, &a, &b);
  for (i = 0; i <= n; i++) {
    x = i;
    if ((n - x) % m == 0)
      y = (n - x) / m;
    else
      y = ((n - x) / m) + 1;
    max = min(max, x * a + y * b);
  }
  printf("%lld\n", max);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
long long int min(long long int m1, long long int m2) {
  if (m1 < m2)
    return m1;
  else
    return m2;
}
int main() {
  long long int n, m, a, b, x, y, max = 1000000, i;
  scanf("%lld %lld %lld %lld", &n, &m, &a, &b);
  for (i = 0; i <= n; i++) {
    x = i;
    if ((n - x) % m == 0)
      y = (n - x) / m;
    else
      y = ((n - x) / m) + 1;
    max = min(max, x * a + y * b);
  }
  printf("%lld\n", max);
  return 0;
}

```