B. Our Tanya is Crying Out Loud

http://codeforces.com/problemset/problem/940/B

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

6

input

Copy

5
5
2
20

output

8

input

Copy

19
3
4
2

output

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.

 

 简单题，但记得特判K=1

// 去吧！皮卡丘! 把AC带回来！
//      へ　　　　　／|
// 　　/＼7　　　 ∠＿/
// 　 /　│　　 ／　／
// 　│　Z ＿,＜　／　　 /`ヽ
// 　│　　　　　ヽ　　 /　　〉
// 　Y　　　　　`　 /　　/
// 　ｲ●　､　●　　⊂⊃〈　　/
// 　()　 へ　　　　|　＼〈
// 　　>ｰ ､_　 ィ　 │ ／／
// 　 / へ　　 /　ﾉ＜| ＼＼
// 　 ヽ_ﾉ　　(_／　 │／／
// 　　 7　　　　　　　|／
// 　　 ＞―r￣￣`ｰ―＿
//**************************************
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define bug printf("***********\n");
#define mp make_pair
#define pb push_back
const int maxn = 3e5 + 10;
const int maxx = 1e6 + 10;
// name*******************************
ll n, k, A, B;
ll ans = 0;
// function******************************

//***************************************
int main() {
  cin >> n >> k >> A >> B;
  if(k==1){
    cout<<(n-1)*A;
    return 0;
  }  
  while (n) {
    if (n < k) {
      ans += (n - 1) * A;
      break;
    }
    int t = n / k;
    ans += (n - k * t) * A + min((k - 1) * A * t, B);
    n = t;
  }
  cout << ans;

  return 0;
}