codeforces 676C

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

Copy

4 2
abba

output

4

input

Copy

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：

给定一个只含字母a和b的字符串，你能够挑出<=k个字母来，让他变化成你想要的字母，问，在此变换下能够得到的最长的连续相同子串的长度是多少？

 分析：

分两种情况，要连续a和连续b。

连续a 时，记录前面有多少个b字母，在这个上面进行尺取就行了。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100005;
char str[maxn];
int numb[maxn];
int numa[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,k;
    scanf("%d%d",&n,&k);
    scanf("%s",str+1);

    for(int i = 1; i <= n; i++) {
        if(str[i]=='b')
            numb[i] = numb[i-1] + 1;
        else numb[i] = numb[i-1];

        if(str[i]=='a')
            numa[i] = numa[i-1] + 1;
        else numa[i] = numa[i-1];
    }

    int L = 1;
    int R = 1;
    int ans = 0;
    int tmp = 0;
    int pos = 0;
    while(R<=n) {
        while(R<=n&&numb[R]-tmp<=k) {
            ans = max(ans,R-pos);
            R++;
        }
        pos++;
        tmp=numb[L];
        L++;
    }

    L = 1;R = 1;
    tmp = 0;
    pos = 0;
    while(R<=n) {
        while(R<=n&&numa[R]-tmp<=k) {
            ans = max(ans,R-pos);
            R++;
        }
        pos++;
        tmp=numa[L];
        L++;
    }

    printf("%d
",ans);


    return 0;
}