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)

#[allow(unused_macros)]
macro_rules! invec {
    ( $t:ty ) => {{
        let mut s = String::new();
        match std::io::stdin().read_line(&mut s) {
            Ok(0) => Vec::<$t>::new(),
            Ok(n) => s.trim().split_whitespace().map(|s| s.parse::<$t>().unwrap()).collect::<Vec<$t>>(),
            Err(_) => Vec::<$t>::new(),
        }
    }}
}

#[allow(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($(
            {
                 splits.next().unwrap().parse::<$t>().unwrap()
            },
        )*)
    }}
}

#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (n, k) = input!(usize, usize);
    let (s,) = input!(String);
    
    let mut divided: Vec<u64> = vec![];

    let mut continued = 0;

    let mut current_div = '2';

    {
        if s.chars().next().unwrap() == '0' {
            divided.push(0);
        }
    }

    for c in s.chars() {
        if c == current_div {
            continued += 1;
        } else {
            if continued != 0 {
                divided.push(continued);
            }
            continued = 1;
            current_div = c;
        }
    }

    divided.push(continued);

    if current_div == '0' {
        divided.push(0);
    }

    let first_right = std::cmp::min(divided.len(), 2 * k + 1);
    
    let mut current_total: u64 = divided[0..first_right].iter().sum();
    let mut max = current_total;
    // println!("{}", current_total);
    // println!("   {}", divided.len() - (2 * k + 1));

    if divided.len() > 2 * k {
        for i in 2..(divided.len() - 2 * k) {
            if i % 2 == 1 {
                continue;
            }
            let left = i;
            let right = left + 2 * k;

            max = std::cmp::max(max, current_total);
            current_total = current_total + divided[right] + divided[right - 1] - divided[left - 1] - divided[left - 2];
            // println!("{} {} : {}", left, right, current_total);
        }
    }
    
    max = std::cmp::max(max, current_total);

    println!("{}", max);
}

fn main() {
    solve();
}