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: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class Main {
	
	static BufferedReader reader;

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		reader = new BufferedReader(new InputStreamReader(System.in));
		
		String[] NKs = readLine().split(" ");
		int N = Integer.parseInt(NKs[0]);
		int K = Integer.parseInt(NKs[1]);
		
		int[] S = readLine().chars().toArray();
		int c_0 = "0".charAt(0);
		int c_1 = "1".charAt(0);
		boolean now = false;
		ArrayList<Integer> cnt_0_1 = new ArrayList<>();
		int tmp = 0;
		for(int i=0;i<N;i++) {
			if(now) {
				if(S[i]==c_0) {cnt_0_1.add(tmp);tmp=1;now=false;}
				else{tmp++;}
			}else {
				if(S[i]==c_0) {tmp++;}
				else{cnt_0_1.add(tmp);tmp=1;now=true;}
			}
		}
		cnt_0_1.add(tmp);
		
		int max_pt = 0;
		int pt=0;
		for(int i=0;i<K*2;i++) {
			if(i<cnt_0_1.size()) {
				pt+=cnt_0_1.get(i);
			}
		}
		max_pt = pt;
		for(int sp=2;sp+K*2-1<=cnt_0_1.size();sp+=2) {
			pt-=cnt_0_1.get(sp-2);
			if(sp-3>=0) {pt-=cnt_0_1.get(sp-3);}
			if(sp+K*2-2<cnt_0_1.size()) {pt+=cnt_0_1.get(sp+K*2-2);}
			if(sp+K*2-1<cnt_0_1.size()) {pt+=cnt_0_1.get(sp+K*2-1);}
			if(pt>max_pt) {max_pt = pt;}
		}
		
		System.out.println(max_pt);
	}

	private static String readLine(){
		try{
			return reader.readLine();
		}catch(Exception e){
			return e.getMessage();
		}
	}
}