Codeforces Round #354 (Div. 2)——C. Vasya and String（尺取）

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".

一开始的渣代码运气好到test65才跪……稍微改下形式就行

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
char s[N];
int main(void)
{
	int n,i,j,cnt,k;
	while (~scanf("%d%d",&n,&k))
	{
		scanf("%s",s);
		int L,R,ma=0,mb=0,c;
		L=R=c=0;
		while (L<n)
		{
			while (R<n&&c<=k)
			{
				if(s[R]=='b')
				{
					if(c==k)
						break;
					c++;
				}	
				R++;
			}
			ma=max(ma,R-L);
			L++;
			c=c-(s[L-1]=='b');
		}
		L=R=c=0;
		while (L<n)
		{
			while (R<n&&c<=k)
			{
				if(s[R]=='a')
				{
					if(c==k)
						break;
					c++;
				}	
				R++;
			}
			mb=max(mb,R-L);
			L++;
			c=c-(s[L-1]=='a');
		}
		printf("%d
",max(ma,mb));
	}
	return 0;
}