C. Vasya and String

原题链接

C. Vasya and String

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”.

大致题意

给你一串只有a和b构成的字符串，字符串长度为n，可修改字符限制个数为k，问能得到的最长连续相同的字符串的长度。非常明显的尺取法的应用。

思路

从左到右扫描，对a和b出现个数计数，只要满足a的个数或者b的个数小于等于k，就把长度加一，否则就将尺子右移（即尺子的开头指针+1，末尾指针+1），如果尺子开头指针的字符为a，则a的计数-1，若果字符是b，则b的计数-1，一直扫描到字符串末尾。

AC代码

#include<iostream>
using namespace std;
char str[100005];
int main(void)
{
    int n,k;
    cin>>n>>k;
    for(int i=0;i<n;i++)
    {
    cin>>str[i];
    }
    int l=0,r=0,ans=0,numa=0,numb=0;
    while(r<n)
    {
        if(str[r]=='a')
        numa++;
        else numb++;
        if(numa<=k||numb<=k)
        {
        ans++;r++;
        }
        else
        {
            if(str[l]=='a')
            numa--;
            else numb--;
            l++;r++;
        }
    }
    cout<<ans<<endl;
    return 0;
}

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