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: Accepted)

use std::cmp;
use std::io;

#[allow(unused_macros)]
macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
}

#[allow(dead_code)]
fn input_line_str() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

#[allow(dead_code)]
fn p<T>(t: T)
where
    T: std::fmt::Display,
{
    println!("{}", t);
}

const SPLIT_DELIMITER: char = ' ';
#[allow(dead_code)]
fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
where
    T: std::str::FromStr,
{
    let mut v: Vec<Vec<T>> = Vec::new();

    for _ in 0..line {
        let vec_line = input_vector();
        v.push(vec_line);
    }
    v
}

#[allow(dead_code)]
fn input_vector<T>() -> Vec<T>
where
    T: std::str::FromStr,
{
    let mut v: Vec<T> = Vec::new();

    let s = input_line_str();
    let split_result = s.split(SPLIT_DELIMITER);
    for z in split_result {
        let buf = match z.parse() {
            Ok(r) => r,
            Err(_) => panic!("Parse Error"),
        };
        v.push(buf);
    }
    v
}

#[allow(dead_code)]
fn str2vec(s: &str) -> Vec<char> {
    let mut v: Vec<char> = Vec::new();
    for c in s.chars() {
        v.push(c);
    }
    v
}

fn main() {
    let n: usize;
    let k: usize;
    input!(n, k);
    let mut s = str2vec(&input_line_str());
    let mut v: Vec<usize> = Vec::new();
    v.push(0);
    let mut prev = '1';
    for z in &s {
        if *z == prev {
            let temp = v.len() - 1;
            v[temp] += 1;
        } else {
            v.push(1);
        }
        prev = *z;
    }
    if *s.last().unwrap() == '0' {
        v.push(0);
    }

    let mut max = 0;
    let mut buf = 0;
    for z in 0..2 * k + 1 {
        if z >= v.len() {
            break;
        }
        buf += v[z];
    }
    max = buf;

    let mut left = 0;
    let mut right = 2 * k;
    loop {
        right += 1;
        if right >= v.len() {
            buf += 0;
        } else {
            buf += v[right];
        }
        right += 1;
        if right >= v.len() {
            buf += 0;
        } else {
            buf += v[right];
        }

        buf -= v[left];
        if left >= v.len() {
            break;
        }
        left += 1;
        if left >= v.len() {
            break;
        }
        buf -= v[left];
        left += 1;
        max = cmp::max(buf, max);
        if right >= v.len() {
            break;
        }
    }
    p(max);
}