#include <bits/stdc++.h>
using namespace std;
void freeopen() {
  freopen(
      ""
      ".in",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
const int D = 1e6 + 1;
const int inf = (int)(1u << 31) - 1u;
const long long INF = (long long)(1ull << 63) - 1ull;
const int N = 1e5 + 1;
const int E = 1e3 + 1;
const int edge = 1e9 + 9LL;
const int mod = 1e9 + 7ll;
const int dx[] = {+1, -1, 0, 0};
const int dy[] = {0, 0, +1, -1};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int n, m, a, b;
  cin >> n >> m >> a >> b;
  if (m * a > b) {
    int ans = 0;
    while (n > 0) {
      n -= m;
      ans += b;
    }
    n += m;
    int num = ans - b;
    while (n > 0) {
      n--;
      num += a;
    }
    if (ans < num)
      cout << ans;
    else
      cout << num;
  } else
    cout << n * a;
  return 0;
}

### Prompt

Generate 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;
void freeopen() {
  freopen(
      ""
      ".in",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
const int D = 1e6 + 1;
const int inf = (int)(1u << 31) - 1u;
const long long INF = (long long)(1ull << 63) - 1ull;
const int N = 1e5 + 1;
const int E = 1e3 + 1;
const int edge = 1e9 + 9LL;
const int mod = 1e9 + 7ll;
const int dx[] = {+1, -1, 0, 0};
const int dy[] = {0, 0, +1, -1};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int n, m, a, b;
  cin >> n >> m >> a >> b;
  if (m * a > b) {
    int ans = 0;
    while (n > 0) {
      n -= m;
      ans += b;
    }
    n += m;
    int num = ans - b;
    while (n > 0) {
      n--;
      num += a;
    }
    if (ans < num)
      cout << ans;
    else
      cout << num;
  } else
    cout << n * a;
  return 0;
}

```