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)

fn solve() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };
    let n: usize = sc.read();
    let k: usize = sc.read();
    let ss: Vec<char> = sc.chars();
    let mut pre = ss[0];
    let mut sum = vec![0; n + 1];

    if pre == '0' {
        sum[1] = 1;
    }
    for i in 2..n + 1 {
        if pre == '1' && ss[i - 1] == '0' {
            sum[i] = 1;
        }
        pre = ss[i - 1];
        sum[i] += sum[i - 1];
    }
    let mut result = 0;
    for i in 1..n + 1 {
        let mut left = i;
        let mut right = n + 10;
        let mut cnt = 0;
        while cnt < 300 {
            let med = (right + left) / 2;
            cnt += 1;
            if med > n {
                right = med;
                continue;
            }

            let tmp = sum[med] - sum[i - 1];

            if tmp <= k {
                result = std::cmp::max(result, med - i + 1);
                left = med;
            } else {
                right = med;
            }
        }
    }
    println!("{}", result);
}

fn main() {
    std::thread::Builder::new()
        .stack_size(64 * 1024 * 1024) // 64MB
        .spawn(|| solve())
        .unwrap()
        .join()
        .unwrap();
}

//snippet from kenkoooo
pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self.stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_line(&mut self) -> String {
        use std::io::Read;
        let buf = self.stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b'\n' || b == b'\r')
            .take_while(|&b| b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}