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: Wrong Answer)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

const INF: i64 = 1 << 60;
#[fastout]
fn main() {
    input! {
    n:usize,k:usize,c:i64,
    s:Chars,
     }

    let mut left = BTreeSet::new();
    let mut right = BTreeSet::new();
    let mut prev = -INF;
    for (i, &now) in s.iter().enumerate() {
        let i = i as i64;
        if left.len() >= k {
            break;
        }
        if now == 'o' {
            if i - prev > c {
                prev = i;
                left.insert(i);
            }
        }
    }
    let mut prev = INF;
    for (i, &now) in s.iter().enumerate().rev() {
        let i = i as i64;
        if right.len() >= k {
            break;
        }
        if now == 'o' {
            if prev - i > c {
                prev = i;
                right.insert(i);
            }
        }
    }
    for l in left.iter() {
        if right.contains(&l) {
            println!("{}", l + 1);
        }
    }
}