Codeforces Round #354 (Div. 2) C

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 denotes beauty 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".

题意：长度为n的字符串 只有‘a’，‘b’ 组成  只能改变k个字符 输出连续相同字符的最大长度

题解：补起来 还是暴力题

        以某一个位置为起点 计算最大值

        之后反转所有的字母 重复fun（）；

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#define ll __int64
#define pi acos(-1.0) 
using namespace std;
int n,k;
char a[100005];
int ans;
void  fun()
{
    int gg=1,exm=0;
    for(int i=1;i<=n;i++)
    {
        while(gg<=n&&(a[gg]=='a'||exm<k))//gg存下 能到达的最远位置，并且以i+1为起点 
        {                                // 最少也能到达 gg
            if(a[gg]=='b')
            exm++;
            gg++;
        }
    //    cout<<gg<<endl;
        ans=max(ans,gg-i);
        if(a[i]=='b') // 除去 对i+1位置为起点的影响
         exm--;
    }
}
int main()
{
     scanf("%d %d",&n,&k);
     getchar();
     for(int i=1;i<=n;i++)
     scanf("%c",&a[i]);
     ans=0;
     fun();
     for(int i=1;i<=n;i++)
     {
         if(a[i]=='a')
         a[i]='b';
         else
         a[i]='a';
     } 
     fun();
     cout<<ans<<endl;
    return 0;
}