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: Wrong Answer)

#include<stdio.h>

int main(void)
{
	int n,k,i,count=0,count2=0,h=0,max=0,tmp;
	char s[100001];
	int a[100001]={0};
	
	scanf("%d%d",&n,&k);
	
	scanf("%s",s);
	
	if(s[0]=='0'){
		a[count2]=0;
		count2++;
		count++;
	}
	else
		count++;
		
	for(i=1;i<n;i++){
		if(s[i-1]==s[i])
			count++;
		else{
			a[count2]=count;
			count2++;
			count=1;
		}
	}
	
	if(count2<=2*k-1){
		printf("%d",n);
		return 0;
	}
	
	for(i=0;i<2*k+1;i++)
		h+=a[i];
	
	max=h;
	
	for(i=0;i*2<=count2-2*k+1;i++){
		h=h-a[i]-a[i+1]+a[2*k+i]+a[2*k+i+1];
		if(h>max)
			max=h;
	}
	
	printf("%d",max);
	return 0;
}