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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var sc = bufio.NewScanner(os.Stdin)

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

func main() {
	buf := make([]byte, 10000)
	sc.Buffer(buf, 1000000)
	nk, s := strings.Split(nextLine(), " "), nextLine()
	k, _ := strconv.Atoi(nk[1])

	// 連続する数を記録
	data := []int{}
	cnt := 0
	currNum := "1"
	hasLast := false
	for _, c := range strings.Split(s, "") {
		if c == currNum {
			cnt++
			hasLast = true
		} else {
			hasLast = false
			data = append(data, cnt)
			cnt = 1
			currNum = c
		}
	}
	if hasLast {
		data = append(data, cnt)
	}
	if len(data)%2 == 0 {
		data = append(data, 0)
	}
	solution := 0
	for i := 0; (i + (2*k + 1)) <= len(data); i += 2 {
		target := data[i:(i + (2*k + 1))]
		total := 0
		for _, v := range target {
			total += v
		}
		if solution < total {
			solution = total
		}
	}
	fmt.Println(solution)
}