codeforces 676C Vasya and String（尺取法）

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 asubstring (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 andk (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 thank characters.

Examples

Input

4 2abba

Output

4

Input

8 1aabaabaa

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".

                

解题思路：尺取法，注意要分为两种情况讨论，都转化为a或者都转化为b。

#include<bits/stdc++.h>using namespace std;typedef long long LL;typedef pair<int,int> PII;#define MP make_pair#define PB push_back#define AA first#define BB second#define OP begin()#define ED end()#define SZ size()#define cmin(x,y) x=min(x,y)#define cmax(x,y) x=max(x,y)const LL MOD = 1000000007;const double PI = acos(-1.);const double eps = 1e-9;char c[100005];int A[100005];int fa(int le,int re){return A[re]-A[le-1];}int fb(int le,int re){return re-le+1-fa(le,re);}int main(){int i,j,k,l,r;int n,m;while(~scanf("%d%d",&n,&m)){scanf("%s",c+1);A[0]=0;for(i=1;i<=n;i++)A[i]=A[i-1]+(c[i]=='a');//保存前n位有几个aint ans=m;//至少可以得到一个长度为m的字串l=1;r=1;while(l<=n)        {            while(r<=n && fa(l,r)<=m)            {                cmax(ans,r-l+1);                r++;            }            l++;        }        l=1;r=1;while(l<=n)        {            while(r<=n && fb(l,r)<=m)            {                cmax(ans,r-l+1);                r++;            }            l++;        }cout<<ans<<"\n";}return 0;}