cf676c Vasya and String 尺取法裸题

C. Vasya and String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotesbeauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.

Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.

The second line contains the string, consisting of letters 'a' and 'b' only.

Output

Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters.

Examples

input

4 2
abba

output

4

input

8 1
aabaabaa

output

5

Note

In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".

In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".

已经记不得是第几次在比赛中见到尺取法了……这个方法虽然说不上是什么高深的算法，但真的很有用……

这个题目尺取起来也很简单，两次尺取，一次刷b，一次刷a，看哪次长就行了。以b为例子。num(b)>k为end指针停止的条件，每次计算end指针停止后的（head-end）长度就行。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int i,j,k,m,n;
	scanf("%d%d",&n,&k);
	getchar();
	char s[100005];
	gets(s);
	int l=strlen(s);
	int head=0;
	int end=0;
	int max1=-1;
	int a=0,b=0;
	 int flag=0; 
	while(!flag)
	{
		while(b<=k  && end<l)
		{
			
			if(s[end]=='a')a++;
			 else b++;
			 end++;
		}
		if(end==l)flag=1;
        if(b>k)
		{end--;
		 b--;
   		}
		max1=max(max1,(end-head));
		while(b>=k && head<=end)
		 {
		 	if(s[head]=='a')a--;
		 	 else b--;
		 	head++;
		 }
		
		
	}
	a=0,b=0;
	flag=0;
	 int max2=-1; 
	 head=0;
	 end=0;
	while(!flag)
	{
		while(a<=k  && end<l)
		{
			
			if(s[end]=='a')a++;
			 else b++;end++;
		}
		if(end==l)flag=1;
		if(a>k)
		{end--;
		 a--;
		}
		max2=max(max2,(end-head));
		if(flag)break;
	//	printf("tou:%d wei:%d:\n",head,end,max2);
		while(a>=k && head<=end)
		 {
		 	if(s[head]=='a')a--;
		 	 else b--;
		 	head++;
		 }
	}
	int Max=max(max1,max2);
	printf("%d\n",Max);
	return 0;
 } 

诶，中间的细节总是会出问题，智商还是太低……