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::io;
use std::str::FromStr;

fn input_tuple() -> (usize, usize) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<usize> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn input_char_vec() -> Vec<char> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().chars().collect()
}

fn main() {
    let (n, k) = input_tuple();
    let s = input_char_vec();
    let mut count: Vec<i32> = Vec::new();
    let mut pre = '2';
    for c in s {
        if pre == c {
            if c == '0' {
                let idx = count.len() - 1;
                count[idx] += 1;
            } else {
                let idx = count.len() - 1;
                count[idx] += -1;
            }
        } else {
            if c == '0' {
                count.push(1);
            } else {
                count.push(-1);
            }
        }
        pre = c; 
    }
    // println!("{:?}", count);
    let mut buf = 0;
    let mut k_count = 0;
    let mut i = 0;
    while k_count < k {
        if i >= count.len() {
            break;
        }
        if count[i] > 0 {
            buf += count[i];
            k_count += 1;
            if i + 1 < count.len() {
                buf -= count[i+1];
            }
            i += 2;
        } else {
            buf -= count[i];
            i+=1;
        }   
    }
    let mut max = buf;
    // println!("{}", max);
    for j in i..count.len() {
        if count[j] < 0 {
            continue;
        } else {
            if j >= 1+2*k {
                buf += count[j-2*k-1];
            }            
            buf -= count[j-2*k];
            buf += count[j];
            if j+1 < count.len() {
                buf -= count[j+1];
            }
            if buf > max {
                max = buf;
            }
        }
        // println!("{}, {}", j, buf);
    }
    println!("{}", max);
}