Codeforces 676C Vasya and String (两点法)

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本文探讨了如何通过更改不多于k个字符来最大化字符串的美丽值，即最长连续相同字符子串的长度。介绍了使用双指针法解决此问题的具体实现。

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

Input

8 1
aabaabaa

Output

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".

题目链接：http://codeforces.com/contest/676/problem/C

题目大意：一个字符串只包含a和b，可以改变其中k个，求最大的美丽值，美丽值定义为子串中所含字符均相同的的最长子串的长度

题目分析：显然对于k次操作，要么全换a要么全换b，对这两种情况取最大即可，算的时候用两点法，r向右直到k用完，然后l从左开始加，不断还原k值

#include <cstdio>
#include <algorithm>
using namespace std;
int const MAX = 1e5 + 5;
char s[MAX];
int n, k;

int cal(char x)
{
    int l = 0, r = 0, ans = 0, cnt = 0;
    while(l < n && r < n)
    {
        while((s[r] == x || cnt < k) && r < n)
        {
            if(s[r] != x)
                cnt ++;
            r ++;
        }
        ans = max(ans, r - l);
        while(s[l] == x && l <= r)
            l ++;
        l ++;
        cnt --;
    }
    return ans;
}

int main()
{
    scanf("%d %d", &n, &k);
    scanf("%s", s);
    printf("%d\n", max(cal('a'), cal('b')));
}