CodeForces - 165B Burning Midnight Oil （二分）

题目链接：http://codeforces.com/problemset/problem/165/B点击打开链接

B. Burning Midnight Oil

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

因为没算一个数是否达到的复杂度是log级别的

因此可以用二分再降一个log

#include <bits/stdc++.h>
using namespace std;
int cal(int v,int k)
{
	int sum=0;
	while(v)
	{
		sum+=v;
		v=v/k;
	}
	return sum;
}
int main()
{
	int n,k;
	cin >>n >>k;
	int mid=n;
	int ans=0;
	int l=0;
	int r=1e9;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(cal(mid,k)>=n)
		{
			ans=mid;
			r=mid-1;
		}
		else
			l=mid+1;
	}
	cout <<ans <<endl;
}