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: Wrong Answer)

use proconio::input;

fn main() {
  input! {
    _n: usize,
    mut k: usize,
    s: String,
  }
  let s = s.chars().collect::<Vec<char>>();
  
  let mut all_one = true;
  for c in s.iter() {
    if c == &'0' {
      all_one = false;
      break
    }
  }
  
  if all_one {
    println!("{}", s.len());
    return
  }
  
  let mut index = 0;
  let mut stack: Vec<usize> = vec![];
  let mut last_start = 0;
  let mut max = 0;
  while index < s.len() {    
    if s[index] == '0' {
      stack.push(index);
      k -= 1;
    }
    
    if k == 0 {
      while index < s.len() && s[index] == '0' {
        index += 1;
      }
      
      while index < s.len() && s[index] == '1' {
        index += 1;
      }
      
      max = std::cmp::max(index - last_start, max);
      last_start = stack.remove(0) + 1;
      k += 1;
    } else {
      index += 1;    
    }
  }
  
  if last_start < s.len() {
    max = std::cmp::max(max, s.len() - 1 - last_start);
  }
  
  println!("{}", max);
}