codeforces 354 div2 C Vasya and String 前缀和

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 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
题意：一个只由'a','b'组成的字符串，长度为n，可以改变k个，求连续最长；
思路：找出连续a的长度，包括0，同样的b也是，求前缀和，求出以k为长度最长的子串；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
#define pi 4*atan(1)
//#pragma comment(linker, "/STACK:102400000,102400000")
int flaga[],jia;
int flagb[],jib;
int disa[];
int disb[];
int suma[];
int sumb[];
char a[];
int main()
{
    int x,y,z,i,t;
    scanf("%d%d",&x,&y);
    scanf("%s",a+);
    flaga[]=;
    jia=;
    flagb[]=;
    jib=;
    for(i=;i<=x;i++)
    {
        if(a[i]=='a')
        flaga[jia++]=i;
        else
        flagb[jib++]=i;
    }
    flaga[jia++]=x+;
    flagb[jib++]=x+;
    for(i=;i<jib;i++)
    disa[i]=flagb[i]-flagb[i-]-;
    for(i=;i<jia;i++)
    disb[i]=flaga[i]-flaga[i-]-;
    for(i=;i<=;i++)
    suma[i]+=suma[i-]+disa[i];
    for(i=;i<=;i++)
    sumb[i]+=sumb[i-]+disb[i];
    int ans=;
    for(i=;i<jia;i++)
    ans=max(ans,sumb[i+y]-sumb[i-]);
    int ans1=;
    for(i=;i<jib;i++)
    ans1=max(ans1,suma[i+y]-suma[i-]);
    if(min(jia,jib)-<=y)
    printf("%d\n",x);
    else
    {
        printf("%d\n",max(ans1,ans)+y);
    }
    return ;
}