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: Time Limit Exceeded)

import java.util.*;
class Group {
    char color;
    int cnt;
    public Group(char color, int cnt){
	this.color = color;
	this.cnt = cnt;
    }
}
class Main {
    public static void main(String[] args) {
	Scanner sc = new Scanner(System.in);
	int N = Integer.parseInt(sc.next());
	int K = Integer.parseInt(sc.next());
	String str = sc.next();
	if(N<=K) {
	    System.out.println(N);
	    return;
	}
	Group[] groups = new Group[N];
	int gcnt = 0;
	for(int i = 0; i<N; i++) {
	    char cur = str.charAt(i);
	    if(i==0) {
		groups[0] = new Group(cur, 1);
		continue;
	    }
	    if(cur == groups[gcnt].color) {
		groups[gcnt].cnt++;
	    }else {
		gcnt++;
		groups[gcnt] = new Group(cur, 1);
	    }
	}
	/*
	for(int i = 0; i<=gcnt; i++) {
	    System.out.print(groups[i].cnt +" ");
	}
	*/
	int max = 0;
	for(int i = 0; i<=gcnt; i++) {
	    // update i
	    int cnt0 = 0;
	    int tmp = 0;
	    int nextI = i;
	    while(cnt0<K && i<=gcnt) {
		tmp += groups[i].cnt;
		if(groups[i].color=='0')cnt0++;
		i++;
	    }
	    if(i<=gcnt) {
		tmp+=groups[i].cnt;
		i--;
	    }
	    i = nextI;
	    //	    System.out.println(tmp);
	    max = Math.max(tmp,max);
	}
	System.out.println(max);
    }
}