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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
int max(int x, int y){return x<y?y:x;}
int main(void){
	int n,k;
	scanf("%d%d",&n,&k);
	char S[100010];
	scanf("%s",S);
	int one[100010][2]={0};
	int c=1;
	one[0][0]=one[0][1]=0;
	if(S[0]=='1'){
		one[1][0]=0;
	}
	for(int i=1;i<n;i++){
		if(S[i-1]=='0' && S[i]=='1'){
			one[c][0] = i+1;
		}
		if(S[i-1]=='1' && S[i]=='0'){
			one[c][1] =i;
			c++;
		}
	}
	for(int i=c;i<c+k;i++){
		one[i][0] = one[i][1] = n+1;
	}
	
	int ans=0,all=1;
	for(int i=0;i<n;i++){
		if(S[i]=='0')all=0;
		ans = max(ans,one[i+k][1]-one[i][0]);
	}
	if(all)ans=n;
	
	printf("%d",ans);
	
	return 0;
}