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 (
    "fmt"
)

const ZERO byte = 48
const ONE byte = 49

type span struct {
    head int
    tail int
}

func (s *span) Len() int {
    return s.tail - s.head
}

func main() {
    var n,k int
    var s string
    fmt.Scan(&n, &k, &s)

    var spans = []span{}
    var current = span{}
    var inSpan = false

    for i:=0; i < len(s); i++ {
        if s[i] == ZERO {
            if inSpan {
                inSpan = false
                current.tail = i
                spans = append(spans, current)
            }
        }

        if s[i] == ONE {
            if !inSpan {
                inSpan = true
                current = span{ head: i }
            }
        }
    }

    if inSpan {
        inSpan = false
        current.tail = len(s)
        spans = append(spans, current)
    }

    var j = 0
    var max = 0
    for {
        if (j+k) > len(spans) - 1 {
            break
        }

        if max < (spans[j+k].tail - spans[j].head) {
            max = (spans[j+k].tail - spans[j].head)
        }
        j++
    }

    fmt.Println(max)
}