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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Chars;
use std::cmp::max;

fn main()
{
    input! {
    n:i32,
    mut k:i32,
    s:Chars
    }

    let mut ans = 0;
    let mut last = 0;
    let mut index = 0;
    while index < n {
        if s[index as usize] == '0'
        {
            k -= 1;
            while index + 1 < n && s[index as usize] == '0' {
                index += 1;
            }

            if k < 0 {
                loop {
                    last += 1;
                    if s[last as usize] == '1' && s[(last - 1) as usize] == '0' {
                        break;
                    }
                }
                k += 1;
            }
        }
        ans = max(ans, index - last + 1);
        index += 1;
    }

    println!("{}", ans);
}