Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in Rust (1.42.0) (judge verdict: Accepted)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	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 c: usize = s.next().unwrap().parse().unwrap();
	let s: Vec<_> = s.next().unwrap().chars().collect();
	let left = {
		let mut r = vec![0; n];
		let mut i = 0;
		let mut cnt = 1;
		while i < n {
			if s[i] == 'o' {
				r[i] = cnt;
				i += c + 1;
				cnt += 1;
			} else {
				i += 1;
			}
		}
		r
	};
	let right = {
		let mut r = vec![0; n];
		let mut cnt = k as i32;
		let mut i = 0;
		while i < n {
			if s[n - 1 - i] == 'o' {
				r[n - 1 - i] = cnt;
				i += c + 1;
				cnt -= 1;
			} else {
				i += 1;
			}
		}
		r
	};

	use std::io::Write;
	let out = std::io::stdout();
	let mut out = std::io::BufWriter::new(out.lock());
	for i in 0..n {
		if s[i] == 'x' {
			continue;
		}
		if left[i] == right[i] && left[i] > 0 {
			writeln!(out, "{}", i + 1).unwrap();
		}
	}
}