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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::str::FromStr;

fn parse() -> (u32, Vec<u32>) {
    let mut line = String::new();

    std::io::stdin().read_line(&mut line).unwrap();
    let k = parse_1(&line);

    line.clear();
    std::io::stdin().read_line(&mut line).unwrap();
    let s = parse_2(&line);

    (k, s)
}

fn parse_1(line: &str) -> u32 {
    let mut it = line.split_whitespace();

    let _n = u32::from_str(it.next().unwrap()).unwrap();
    let k = u32::from_str(it.next().unwrap()).unwrap();

    k
}

fn parse_2(line: &str) -> Vec<u32> {
    let mut s = Vec::with_capacity(line.bytes().len());

    let mut prev = '1' as u8;
    let mut count = 0;

    for c in (line).bytes() {
        if c == prev {
            count = count + 1;
        } else {
            s.push(count);
            count = 1;
            prev = c;
        }
    }
    s.push(count);

    s
}

fn count(k: u32, s: &[u32]) -> u32 {
    let end = (2 * k + 1) as usize;
    if s.len() <= end {
        return s.iter().sum();
    }

    let mut ret = s[0..end].iter().sum();
    let mut current = ret;

    let mut it0 = s.iter();
    let mut it1 = s.iter().skip(end);

    loop {
        if let Some(n) = it1.next() {
            current = current + n;
        } else {
            break;
        }

        if let Some(n) = it1.next() {
            current = current + n;
        } else {
            break;
        }

        current = current - it0.next().unwrap();
        current = current - it0.next().unwrap();

        if ret < current {
            ret = current;
        }
    }

    ret
}

fn main() {
    let (k, s) = parse();
    println!("{}", count(k, &s));
}

#[test]
fn sample_1() {
    let k = 1;
    let s = parse_2("00010");
    assert_eq!(&s, &[0, 3, 1, 1]);
    assert_eq!(count(k, &s), 4);
}

#[test]
fn sample_2() {
    let k = 2;
    let s = parse_2("11101010110011");
    assert_eq!(&s, &[3, 1, 1, 1, 1, 1, 2, 2, 2]);
    assert_eq!(count(k, &s), 8);
}

#[test]
fn sample_3() {
    let k = 1;
    let s = parse_2("1");
    assert_eq!(&s, &[1]);
    assert_eq!(count(k, &s), 1);
}