Burning Midnight Oil CodeForces - 165B

Burning Midnight Oil

CodeForces - 165B

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.

Example

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.

这个题用二分，只要是遇到可以从头到尾遍历暴力检查的题目，那就可以用二分。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
int n,k;
int q_pow(int a,int b){//快速幂，这道题用不用都一样
int ans = 1;
while(b){
if(b&1){
ans *= a;
}
b >>= 1;
a *= a;
}
return ans;
}
int check(int v){//检查是否符合题意
int i = 0;
int sum = 0;
while(v/q_pow(k,i)){
sum += (v/q_pow(k,i));
i++;
if(sum >= n)return 1;
}
return 0;
}
int main(){
scanf("%d%d",&n,&k);
int l = 1,r = n;
while(l <= r){//二分查找
int mid = (l+r)/2;
if(check(mid)){
r = mid-1;
}
else{
l = mid+1;
}
}
printf("%d\n",l);
return 0;
}