Greedy Arkady(思维)

$k$ people want to split $n$

 candies between them. Each candy should be given to exactly one of them or be thrown away.

The people are numbered from $1$

 to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$

) will be thrown away.

Arkady can't choose $x$

 greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$

 times, as it is considered a slow splitting.

Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$

.

Input

The only line contains four integers $n$

, $k$, $M$ and $D$ ($2\le n\le {10}^{18}$, $2\le k\le n$, $1\le M\le n$, $1\le D\le min(n,1000)$, $M\cdot D\cdot k\ge n$

) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.

Output

Print a single integer — the maximum possible number of candies Arkady can give to himself.

Note that it is always possible to choose some valid $x$

.

Examples

input

Copy

20 4 5 2

output

Copy

8

input

Copy

30 9 4 1

output

Copy

4

Note

In the first example Arkady should choose $x=4$

. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$

 candies in total.

Note that if Arkady chooses $x=5$

, he will receive only $5$ candies, and if he chooses $x=3$, he will receive only $3+3=6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x=1$nor $x=2$ because in these cases he will receive candies more than $2$

 times.

In the second example Arkady has to choose $x=4$

, because any smaller value leads to him receiving candies more than $1$

 time

.

题意：k个人分n个糖果，第一个人先拿x个，然后每个人轮流拿x个，直至不到x个糖果（可以循环轮流）。

    问第一个人可分的的最多的糖果的个数。

一开始想的用二分，但是精度不好控制，条件不好写。后纯思维，采取暴力。第一个人要想分最多，其他人分

i-1次，第一个人就分i次，才能够保证糖数最多。

代码：

#include<stdio.h>
#include<string.h>
#define ll long long int
int main()
{
   ll n,k,m,d,i,x,countt;

   scanf("%lld %lld %lld %lld",&n,&k,&m,&d);

   countt=0;

   for(i=1;i<=d;i++){

      x=n/(k*(i-1)+1);  //其他人都是i-1次，只有第一个是i次，所以总共的是k*(i-1)+1次；

      if(x==0)
        break;

      if(x>m)
       x=m;

      countt=countt>i*x?countt:i*x;

   }

   printf("%lld\n",countt);

   return 0;
}