cf466A Cheap Travel

A. Cheap Travel

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

题意是要买n张票，可以买1张价格是a，也可以买m张价格是b

忘记考虑全买m张的最后剩下来直接买m张比一张一张更优的情况了……wa了两次

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define LL long long
using namespace std;
LL n,m,a,b;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int main()
{
	n=read();m=read();a=read();b=read();
	printf("%d",min(min(n*a,(n/m)*b+(n%m)*a),(n/m+1)*b));
}