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::io::stdin;

fn main() {
    let k: Vec<usize> = read().split(" ").map(|n| n.parse().unwrap()).collect();
    let mut s = read();
    let n = k[1];
    for _ in 0..n {
        let count = get_count('0', &s);
        if count > 0 {
            let b = std::iter::repeat('0').take(count).collect::<String>();
            let r = std::iter::repeat('1').take(count).collect::<String>();
            {
                let t = s.clone();
                let vec: Vec<&str> = t.split(&b).collect();
                if vec.len() == 2 {
                    s = format!("{}{}{}", vec[0], r, vec[1]);
                } else {
                    let mut max = 0;
                    let mut index = 0;
                    for i in 0..vec.len()-1 {
                        let q = vec[i].len() + vec[i + 1].len();
                        if max < q {
                            max = q;
                            index = i;
                        }
                    }
                    let b = std::iter::repeat('0').take(count).collect::<String>();
                    let b = format!("{}{}{}", vec[index], b, vec[index + 1]);
                    let r = std::iter::repeat('1').take(max + count).collect::<String>();
                    s = s.replace(&b, &r);
                }
            }
        }
    }
    let count = get_count('1', &s);
    println!("{}", count);
}

fn get_count(c: char, t: &str) -> usize {
    let mut cc = String::new();
    loop {
        cc.push(c);
        if t.contains(&cc) {
        } else {
            cc.pop();
            break;
        }
    }
    return cc.len();
}

fn read() -> String {
    let mut s = String::new();
    stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}