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: Accepted)

package main

import (
	"fmt"
)

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

	var s string
	fmt.Scan(&s)

	max, count, right := 0, 0, 0
	for left := 0; left < n; {
		// move right until 0
		for right < n {
			if s[right:right+1] == "0" && count == k {
				break
			}

			if s[right:right+1] == "0" {
				for right < n && s[right:right+1] == "0" {
					right++
				}
				count++
			} else {
				right++
			}
		}

		if max < right - left {
			max = right - left
		}

		// move left over 1st 0
		for left < n && s[left:left+1] != "0" {
			left++
		}
		for left < n && s[left:left+1] == "0" {
			left++
		}
		count--
	}

	fmt.Println(max)
}