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>
#include <string.h>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define FOR(i,m,n) for(i=m;i<n;i++)

int main(void)
{
	long long a,b,i,d=1,e[1000010]={},f[1000010]={},g=0,j,k,l[1000010]={},n[1000010]={},m=0,o;
	char c[1000010];
	scanf("%lld %lld",&a,&b);
	scanf("%s",c);
	for(i=0;i<a;i++){
		if(d==1){
			if(c[i]=='1'){
				e[g]++;
			}else{
				f[g]++;
				d=0;
//              g++;
//              printf("%lld\n",g);
			}
		}else{
			if(c[i]=='0'){
				f[g]++;
			}else{
				g=g+1;
//              	printf("%lld\n",g);
				e[g]++;
				d=1;
			}
		}
	}
	for(i=1;i<1000010;i++){
		l[i]=l[i-1]+e[i-1];
		n[i]=n[i-1]+f[i-1];
		if(f[i-4]==0){
          if(i>4){
				o=i;
				break;
          }
		}
	}
	for(i=0;i<o+3;i++){
		m=max(m,l[i+b+1]-l[i]+n[i+b]-n[i]);
	}
	printf("%lld",m);
	return 0;
}