Our Tanya is Crying Out Loud

Posted 2021-03-06 dealer

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:

1. Subtract 1 from x. This operation costs you A coins.
2. 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.

Examples

input

Copy

9
2
3
1

output

Copy

6

input

Copy

5
5
2
20

output

Copy

8

input

Copy

19
3
4
2

output

Copy

12

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.

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
//#include <xfunctional>
#define ll long long
#define mod 1000000007
using namespace std;
int dir[4][2] = { { 0,1 },{ 0,-1 },{ -1,0 },{ 1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;

int main()
{
    ll n, k, A, B;
    ll res=0;
    cin >> n >> k >> A >> B;
    if (k == 1)
        cout << (n - 1)*A;
    else
    {
        while (n != 1)
        {
            if (n < k)
            {
                res += (n - 1)*A;
                n = 1;
            }
            else
            {
                if (n%k > 0)
                {
                    res += (n%k)*A;
                    n -= n%k;
                }
                if ((n - n / k)*A > B)
                {
                    res += B;
                    n /= k;
                }
                else
                {
                    res += (n - n / k)*A;
                    n /= k;
                }
            }
        }
        cout << res;
    }
    return 0;
}