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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		String S = sc.next();
		StringBuilder S2 = new StringBuilder(S);
		int div=0;
		int temp=0;
		List<Integer> list = new ArrayList<Integer>();
		for (int i=0;i<N;i++) {
			if (S.charAt(i)=='0') {
				if (temp==0) {
					div++;
					list.add(i);
				}
				temp=1;
			} else {
				temp=0;
			}
		}

		int max=0;
		int max_temp=0;
		int k_init=0;
		if (K>=div) {
			max=N;
		}
		for (int i=0;i<N;i++) {
			if (S.charAt(i)=='0') {
				break;
			}
			if (i==N-1) {
				max=N;
			}
		}
		if (max!=N) {
			for (int i=0;i<=div-K;i++) {
				int A=list.get(K-i)-list.get(i);
				max=Math.max(A,max);
			//StringBuilder S1 = new StringBuilder(S2);
				//for (int j=0;j<K;j++) {
					/*
					for (int k=k_init;k<N;k++) {
						if (S1.charAt(k)=='0') {
							S1.setCharAt(k,'1');
						} else if (S.charAt(k)=='1') {
							if (k==0) {
							} else if (S.charAt(k-1)=='0') {
							k_init=k+1;
							break;
							}
						}
					}
					int temp1=0;
					for (int q=0;q<N;q++) {
						if (S1.charAt(q)=='1') {
							temp1++;
							max_temp=Math.max(temp1,max_temp);
						} else if(S1.charAt(q)=='0') {
							temp1=0;
						}
					}
					max=Math.max(max,max_temp);
					*/
				//}
			}
		}
		System.out.println(max);
	}
}