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: Time Limit Exceeded)

use std::io::stdin;

fn main() {
    let mut nk = String::new();
    stdin().read_line(&mut nk).unwrap();
    let nk: Vec<usize> = nk.trim().split_whitespace().flat_map(str::parse).collect();
    let mut s = String::new();
    stdin().read_line(&mut s).unwrap();

    let (n, k) = (nk[0], nk[1]);
    let s = s.trim();
    let mut start = None;
    let mut ps = Vec::new();

    for i in 0..n+1 {
        match s.chars().nth(i) {
            Some('0') if start == None => {
                start = Some(i);
            },
            Some('1') | None if start != None => {
                ps.push((start.unwrap(), i));
                start = None;
            },
            _ => {},
        }
    }
//    println!("{:?}", ps);
    let mut result = None;


    for i in 0..if ps.len() as isize -k as isize + 1 <= 0 {1} else {ps.len()-k+1} {
//        println!("{} {}", i, i + k - 1);
        // ゼロをフリップしたときの連続して逆立ちしている人の数を数える
        let start = if i > 0 {
            ps[i-1].1
        } else {
            0
        };
        let end = if i + k < ps.len() {
            ps[i + k].0
        } else {
            n
        };
        if result == None || result < Some(end - start) {
            result = Some(end - start);
        }
//        println!("{} - {} = {}", end, start, end - start);
    }
    println!("{}", result.unwrap());
}

/*
    let mut ps = Vec::new();
    let mut c = 0;
    let mut d = 0;

    for i in 0..n+1 {
        match s.chars().nth(i) {
            Some('1') => {
                c += 1;
                if d > 0 {
                    ps.push((d, 0));
                    d = 0;
                }
            },
            Some('0') => {
                d += 1;
                if c > 0 {
                    ps.push((c, 1));
                    c = 0;
                }
            },
            None => {
                if c > 0 {
                    ps.push((c, 1));
                }
                if d > 0 {
                    ps.push((d, 0));
                }
            }
            _ => {},
        }
    }
    println!("{:?}", ps);
*/