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)

use proconio::marker::Chars;
use proconio::{fastout, input};
use std::cmp::max;

#[fastout]
fn main() {
    input! {in_n: usize, in_k: i32, in_c: i32, in_s: Chars};
    let mut l_works = vec![0i32; in_n + 1];
    let mut r_works = vec![0i32; in_n + 1];
    let mut l_count = 0;
    let mut r_count = 0;
    let mut l_break = 0;
    let mut r_break = 0;

    for i in 0..in_s.len() {
        if l_break == 0 && in_s[i] == 'o' {
            l_count += 1;
            l_break = in_c + 1;
        }

        if r_break == 0 && in_s[in_n - i - 1] == 'o' {
            r_count += 1;
            r_break = in_c + 1;
        }

        l_works[i + 1] = l_count;
        r_works[in_n - i - 1] = r_count;

        l_break = max(l_break - 1, 0);
        r_break = max(r_break - 1, 0);
    }

    for i in 0..in_n {
        if l_works[i] + r_works[i + 1] < in_k {
            println!("{}", i + 1);
        }
    }
}