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;
        int l=0,r=0;
        int i;
        int len=0;
        char s[1000000];
        scanf("%d%d",&n,&k);
        scanf("%s",s);

        for(i=0;i<k+(s[0]=='1');i++){
                if (r<n-1) r++;
                while(r<n-1&&(s[r]!='1'||s[r+1]!='0')){
                        r++;
                }
        }
        len=r-l+1;
        //printf("%d\n",r);
        while(l<n-1&&r<n){
                if(r<n-1){
                    l++; r++;
                }
                while(l<n-1&&(s[l]!='0'||s[l+1]!='1')) l++;
                if(s[n-1]=='1'){
                    while(r<n-1&&(s[r]!='1'||s[r+1]!='0')) r++;
                }else{
                    while(r<n&&(s[r]!='1'||s[r+1]!='0')) r++;
                }
                if(r-l>len) len=r-l;
                //printf("%d %d\n",l,r);
        }
        printf("%d\n",len);
        return 0;
}