Codeforces Round #476 (Div. 2) [Thanks, Telegram!] C. Greedy Arkady

C. Greedy Arkady

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

$k$ 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

.

题解：
因为d在1000之内，可以枚举d,对于每个d，可以把
n划分为(d-1)*k~d*k块，二分查找即可。
代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    ll n,k,m,d;
    scanf("%lld%lld%lld%lld",&n,&k,&m,&d);
    ll ans=0;
    for(ll t=1;t<=d&&t*k<=n;t++)
    {
        ll l=(t-1)*k,r=t*k;
        while(r-l>1)
        {
            ll mid=(r+l)/2;
            if(n/(mid+1)<m)r=mid;
            else l=mid;
        }
        ans=max(ans,min(n/r*t,t*m));
    }
    printf("%lld\n",ans);
    return 0;
}