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 (1.42.0) (judge verdict: Accepted)

use std::io::Read;
use std::collections::VecDeque;
use std::cmp::max;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

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

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let k: usize = iterator.next().unwrap().parse().unwrap();
    let s: Vec<char> = iterator.next().unwrap().chars().collect();

    let mut q1: VecDeque<(char, usize)> = VecDeque::new();
    let mut current = (s[0], 1);
    for i in 1..n {
        if s[i] != current.0 {
            q1.push_back(current);
            current = (s[i], 1);
        } else {
            current.1 += 1;
        }
    }
    q1.push_back(current);

    let mut count: usize = 0;
    let mut score: usize = 0;
    let mut q2: VecDeque<(char, usize)> = VecDeque::new();
    while count < k && !q1.is_empty() {
        let current = q1.pop_front().unwrap();
        if current.0 == '0' {
            count += 1;
        }
        score += current.1;
        q2.push_back(current);
    }
    if !q1.is_empty() {
        let current = q1.pop_front().unwrap();
        score += current.1;
        q2.push_back(current);
    }

    let mut ans = score;
    while !q1.is_empty() {
        if let Some(removed) = q2.pop_front() {
            score -= removed.1;
            if removed.0 == '1' {
                if let Some(removed2) = q2.pop_front() {
                    score -= removed2.1;
                }
            }
        }
        let next = q1.pop_front().unwrap();
        score += next.1;
        q2.push_back(next);
        if !q1.is_empty() {
            let next = q1.pop_front().unwrap();
            score += next.1;
            q2.push_back(next);
        }
        ans = max(ans, score);
    }
    ans.to_string()
}