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.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.function.Consumer;

import static java.util.stream.Collectors.toList;

public class Main {
    private static final Scanner scanner = new Scanner(System.in);
    private static final Consumer<List<String>> consumer = solve();

    public static void main(String[] args) {
        consumer.accept(readInput());
    }

    private static List<String> readInput() {
        final List<String> lineList = new ArrayList<>();
        while (scanner.hasNextLine()) {
            lineList.add(scanner.nextLine());
        }
        return lineList;
    }

    private static Consumer<List<String>> solve() {
        return args -> {
            final List<Integer> NK = Arrays.stream(args.get(0).split(" ")).map(Integer::valueOf).collect(toList());
            final Integer K = NK.get(1);
            final char[] numList = args.get(1).toCharArray();

            Integer max = 0;
            Integer cnt = 0;
            Integer cntLeft1 = 0;
            Integer remainingK = K;
            List<Integer> stack = new ArrayList<>();
            int index = 0;
            while (index < numList.length) {
                if (numList[index] == '1') {
                    cnt++;
                    cntLeft1++;
                    index++;
                } else {
                    Integer cnt0 = 0;
                    while (index < numList.length && numList[index] == '0') {
                        cnt0++;
                        cnt++;
                        index++;
                    }
                    stack.add(cntLeft1 + cnt0);
                    cntLeft1 = 0;
                    if (remainingK > 0) {
                        remainingK--;
                    } else {
                        cnt -= stack.remove(0);
                    }
                }
                max = Math.max(max, cnt);
            }

            System.out.println(max);
        };
    }

}
