B

One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of n lines of code. Vasya is already exhausted, so he works like that: first he writes v lines of code, drinks a cup of tea, then he writes as much as  lines, drinks another cup of tea, then he writes  lines and so on: , , , ...

The expression  is regarded as the integral part from dividing number a by number b.

The moment the current value  equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished.

Vasya is wondering, what minimum allowable value v can take to let him write not less than n lines of code before he falls asleep.

Input

The input consists of two integers n and k, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1 ≤ n ≤ 10⁹, 2 ≤ k ≤ 10.

Output

Print the only integer — the minimum value of v that lets Vasya write the program in one night.

Examples

Input

7 2

Output

4

Input

59 9

Output

54

Note

In the first sample the answer is v = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task.

In the second sample the answer is v = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than n = 59.

题解：找到一个合适的值V，

【一天，一个非常重要的任务被委托给Vasya--在一个晚上写一个程序。该程序由n行代码组成。Vasya已经耗尽，所以，他的作品就像是：第一，他写道v行代码，喝一杯茶，然后他尽可能多写为线，饮料一杯茶，然后他写道线等：，，， ...

该表达式被视为从数字a除以数字b组成的组成部分。

当前值等于0 的那一刻，Vasya立即睡着了，他只在早上才醒来，当时程序应该已经完成​​。

Vasya很纳闷，有什么最小允许值v可以让他写不低于比ň行代码，他睡着了。】

【题析】：需要用二分查找 

#include<bits/stdc++.h>
using namespace std;

int main() {
    long long  n;
    int m;
    scanf("%lld %d",&n,&m);
    long long v;
    int flag = 0;
    long long  vv;
    long long mid;
    long long left  = 0;
    long long right = n; 
    long long ans = n;
    while(left <= right) {
        mid = (left + right) / 2;
        //printf("%lld %lld %lld
",left,right,mid);
        long long  sum = mid;
        int count = 0;
        long long v = m;
        int vv = mid/v;
        while(vv){        
            sum += vv;
            count++;
            v = v*m;
            vv = mid/v;
        }
        if(flag == 1) break;
        if(sum >= n) {
            
            ans = min(ans,mid);
            right = mid - 1;
            ans = min(ans,mid);
        } else {
            left = mid + 1;
        }
    }
    printf("%lld
",ans);
    return 0;
}