Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>


int main(void){
	char s[100024];
	int str[100024]={0};
	int str_count=0;
	int ans = 0;
	int n,k;
	scanf("%d %d",&n,&k);
	scanf("%s",s);
	int i,j;

	int hoge=1;
	for(i=1;i<n;i++){
		if(s[i]==s[i-1]){
			hoge++;
		}else{
			str[str_count++]=hoge;
			hoge=1;
		}
	}
	str[str_count++]=hoge;
	
	int flag = (s[0] == '0')?0:1;
	for(i=0;i<str_count;i++){
		int piyo=0;
		if(flag==0){
			for(j=i;j<i+k*2;j++){
				piyo+=str[j];
				if(str[j]==0)break;
			}
			flag=1;
		}else{
			for(j=i;j<i+1+k*2;j++){
				piyo+=str[j];
				if(str[j]==0)break;
			}
			flag=0;
		}
		if(ans<piyo){
			ans=piyo;
		}
	}
	printf("%d\n",ans);


	return 0;
}