CodeForces - 940B（模拟）

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本文探讨了一个数学问题，即如何通过两种操作（减一或除以k）将一个数n转换为1，同时使得消耗的硬币数量最少。文章提供了一种有效的算法实现，并通过实例解释了算法的工作原理。

Right now she actually isn't. But she will be, if you don't solve this problem.

You are given integers n, k, A and B. There is a number x, which is initially equal to n. You are allowed to perform two types of operations:

Subtract 1 from x. This operation costs you A coins.
Divide x by k. Can be performed only if x is divisible by k. This operation costs you B coins.

What is the minimum amount of coins you have to pay to make x equal to 1?

Input

The first line contains a single integer n (1 ≤ n ≤ 2·10⁹).

The second line contains a single integer k (1 ≤ k ≤ 2·10⁹).

The third line contains a single integer A (1 ≤ A ≤ 2·10⁹).

The fourth line contains a single integer B (1 ≤ B ≤ 2·10⁹).

Output

Output a single integer — the minimum amount of coins you have to pay to make x equal to 1.

Example

Input

Output

Input

Output

Input

Output

Note

In the first testcase, the optimal strategy is as follows:

Subtract 1 from x (9 → 8) paying 3 coins.
Divide x by 2 (8 → 4) paying 1 coin.
Divide x by 2 (4 → 2) paying 1 coin.
Divide x by 2 (2 → 1) paying 1 coin.

The total cost is 6 coins.

In the second test case the optimal strategy is to subtract 1 from x 4 times paying 8 coins in total.

题意：将一个数n进行两种操作，一种是减一，这会消耗a个硬币，另一种是除以k（如果n可以整除k的话），这会消耗b个硬币。问使n变成1最少消耗多少个硬币？

解：这题难就难在要把情况考虑清楚，比赛的时候先想的是用广搜跑一遍，但是超时了，后来想到了正解，但是因为没有特判k=1的情况一直没有过，最后就放弃了，其实当时离正确答案就差一点点了，很遗憾，没有把情况考虑清楚。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<math.h>
#include<iostream>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#define inf 0x3fffffff
using namespace std;
typedef long long LL;
int n,k,b;
LL a,ans;
int main()
{
    while(~scanf("%d%d%lld%d",&n,&k,&a,&b))
    {
        ans=0;
        if(k==1||n<k)//当n<k或者k=1时只能进行减一操作
        {
            printf("%lld\n",(n-1)*a);
            continue;
        }
        while(n>1)
        {
            if(n%k)//n不能整除k时只能减一
            {
                LL t=n%k;
                if(n<k) t--;//不能减到0，只能减到1
                n-=t;
                ans+=a*t;
            }
            else//若可以整除那就要判断减一操作和整除操作哪个花的硬币更少
            {
                if((n-n/k)*a<b) ans+=(n-n/k)*a;
                else ans+=b;
                n/=k;
            }
        }
        printf("%lld\n",ans);
    }
}