#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, s = 0, i, c = 0, count = 0;
  float a, b, d1, d2;
  cin >> n >> m >> a >> b;
  d1 = 1 / a;
  d2 = m / b;
  if (d1 >= d2) {
    for (i = 1; i <= n; i++) {
      s = s + a;
    }
    cout << s;
  } else {
    for (i = 1; i <= n; i++) {
      if (c >= n)
        break;
      else {
        s = s + b;
        c = c + m;
        count++;
      }
    }
    if (c == n)
      cout << s;
    else {
      if (s < ((s - b) + (n - c + m) * a))
        cout << s;
      else
        cout << (s - b) + (n - c + m) * a;
    }
  }
}

### 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 n, m, s = 0, i, c = 0, count = 0;
  float a, b, d1, d2;
  cin >> n >> m >> a >> b;
  d1 = 1 / a;
  d2 = m / b;
  if (d1 >= d2) {
    for (i = 1; i <= n; i++) {
      s = s + a;
    }
    cout << s;
  } else {
    for (i = 1; i <= n; i++) {
      if (c >= n)
        break;
      else {
        s = s + b;
        c = c + m;
        count++;
      }
    }
    if (c == n)
      cout << s;
    else {
      if (s < ((s - b) + (n - c + m) * a))
        cout << s;
      else
        cout << (s - b) + (n - c + m) * a;
    }
  }
}

```