#include <bits/stdc++.h>
using namespace std;
vector<long long int> prime(1000001, 1);
void is_prime() {
  prime[0] = 0;
  prime[1] = 0;
  for (long long int i = 2; i * i < 1000001; ++i) {
    if (prime[i]) {
      for (long long int j = i * i; j < 1000001; j += i) {
        prime[j] = 0;
      }
    }
  }
}
int main() {
  int i, t = 1;
  while (t--) {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    int ans = 0;
    if (a * m > b) {
      if (n < m) {
        if (a * n < b) {
          ans = a * n;
        } else
          ans = b;
      } else {
        int d = n / m + 1;
        ans = ((d - 1) * b) + (a * (n % m));
        if (d * b < ans) ans = d * b;
      }
    } else {
      ans = a * n;
    }
    cout << ans;
  }
}

### Prompt

Create a solution in Cpp for 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;
vector<long long int> prime(1000001, 1);
void is_prime() {
  prime[0] = 0;
  prime[1] = 0;
  for (long long int i = 2; i * i < 1000001; ++i) {
    if (prime[i]) {
      for (long long int j = i * i; j < 1000001; j += i) {
        prime[j] = 0;
      }
    }
  }
}
int main() {
  int i, t = 1;
  while (t--) {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    int ans = 0;
    if (a * m > b) {
      if (n < m) {
        if (a * n < b) {
          ans = a * n;
        } else
          ans = b;
      } else {
        int d = n / m + 1;
        ans = ((d - 1) * b) + (a * (n % m));
        if (d * b < ans) ans = d * b;
      }
    } else {
      ans = a * n;
    }
    cout << ans;
  }
}

```