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::*;
use std::str::FromStr;
use utils::*;

pub fn main() {
    let i = stdin();
    let mut o = Vec::new();
    run(i.lock(), &mut o);
    stdout().write_all(&o).unwrap();
}

fn run<R: BufRead, W: Write>(i: R, o: &mut W) {
    let mut i = CpReader::new(i);
    let (_n, k) = i.read2::<usize, usize>();
    let s = i.read_line().as_bytes();
    writeln!(o, "{}", solve(s, k)).unwrap();
}
fn solve(s: &[u8], k: usize) -> usize {
    let mut cs = Vec::new();

    let mut i = 0;
    while s.get(i) == Some(&b'1') {
        i += 1;
    }
    cs.push(i);

    while i < s.len() {
        let i_begin = i;
        while s.get(i) == Some(&b'0') {
            i += 1;
        }
        cs.push(i - i_begin);

        let i_begin = i;
        while s.get(i) == Some(&b'1') {
            i += 1;
        }
        cs.push(i - i_begin);
    }

    let mut current = cs[0];
    let mut result = current;
    for i in 0..cs.len() / 2 {
        let j = i * 2 + 1;
        current += cs[j] + cs[j + 1];
        if i >= k {
            let j = (i - k) * 2 + 1;
            current -= cs[j - 1] + cs[j];
        }
        result = max(result, current);
    }
    result
}

mod utils {
    use super::*;

    pub struct CpReader<R: BufRead> {
        r: R,
        s: String,
    }

    macro_rules! fn_read {
        { $f:ident($($t:ident),*) } => {
            #[allow(dead_code)]
            pub fn $f<$($t: FromStr),*>(&mut self) -> ($($t),*) {
                let i = &mut self.read_line().split(' ');
                ($(read_next::<$t,_>(i)),*)
            }
        };
    }

    impl<R: BufRead> CpReader<R> {
        pub fn new(r: R) -> Self {
            CpReader {
                r: r,
                s: String::new(),
            }
        }
        pub fn read_line(&mut self) -> &str {
            self.s.clear();
            self.r.read_line(&mut self.s).unwrap();
            self.s.trim()
        }

        fn_read! { read2(T1, T2) }
        fn_read! { read3(T1, T2, T3) }
        fn_read! { read4(T1, T2, T3, T4) }
        fn_read! { read5(T1, T2, T3, T4, T5) }
    }
    fn read_next<'a, T: FromStr, I: Iterator<Item = &'a str>>(i: &mut I) -> T {
        i.next().unwrap().parse().ok().unwrap()
    }

}