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.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = "1" + sc.next();

		int[] ruiseki = new int[s.length() + 1];

		// 累積和配列を作成
		char preChar = s.charAt(0);
		int gIndex = 1;
		int gNum = 0;
		for (int i = 1; i <= s.length(); i++) {
			gNum++;
			ruiseki[gIndex] = gNum;
			if (i < s.length() && preChar != s.charAt(i)) {
				gIndex++;
				preChar = s.charAt(i);
			}
		}

//		System.out.println(ruiseki);
		int max = gIndex == 1 ? n : 0;
		for (int i = 2; i <= gIndex; i += 2) {
//			System.out.println(2 * (k - 1) + i + 1 + ":index");
//			System.out.println(Math.min(2 * (k - 1) + i + 1, gIndex));
//			System.out.println(ruiseki[Math.min(2 * (k - 1) + i + 1, gIndex)] - ruiseki[i - 2] + ":max?");
			if (max < ruiseki[Math.min(2 * (k - 1) + i + 1, gIndex)] - ruiseki[i - 2]) {
				max = ruiseki[Math.min(2 * (k - 1) + i + 1, gIndex)] - ruiseki[i - 2];
				if (i == 2) max--;
			}
		}
		System.out.println(max);
	}
}