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)

fn main() {
	use std::io::Read;
	let mut s = String::new();
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let k: usize = s.next().unwrap().parse().unwrap();
	let s: Vec<char> = s.next().unwrap().chars().collect();
	let mut l = 0;
	let mut r = 0;
	let mut cnt = 0;
	let mut ans = 0;
	while l < s.len() {
		while r < s.len() {
			if s[r] == '0' {
				if cnt >= k {
					break;
				}
				while r < s.len() && s[r] == '0' {
					r += 1;
				}
				cnt += 1;
			} else {
				r += 1;
			}
		}
		if r - l > ans {
			ans = r - l;
		}
		if s[l] == '1' {
			while l < s.len() && s[l] == '1' {
				l += 1;
			}
		}
		if l < s.len() {
			if cnt > 0 {
				cnt -= 1;
			}
			while l < s.len() && s[l] == '0' {
				l += 1;
			}
		}
	}
	println!("{}", ans);
}