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::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default, PartialEq, Eq, PartialOrd, Ord)]
struct Struct;

//#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        k: usize,
        c: i32,
        s: Bytes,
    }

    /*
        d0[i] = (j,k) : today + i 日目以前に、j日働ける。最後に働いたのは today+k 日目
        d1[i] = (j,k) : today + i 日目以降に、j日働ける。次に働くのは、遅くとも today+k日目
    */

    let mut d0 = vec![(0usize, 0); n];
    let mut d1 = vec![(0usize, 0); n];

    {
        let mut last_work_i = -c - 1;
        let mut ct = 0;
        for i in 0..n {
            if last_work_i + c < i as i32 && s[i] == b'o' {
                ct += 1;
                last_work_i = i as i32;
            }
            d0[i] = (ct, last_work_i);
        }
    }

    {
        let mut last_work_i = n as i32 + c;
        let mut ct = 0;
        for i in (0..n).rev() {
            if i as i32 + c < last_work_i && s[i] == b'o' {
                ct += 1;
                last_work_i = i as i32;
            }
            d1[i] = (ct, last_work_i);
        }
    }

    for i in 0..n {
        let (work0, last_work) = if i == 0 { (0, -c - 1) } else { d0[i - 1] };
        let (work1, next_work) = if i == n - 1 {
            (0, n as i32 + c)
        } else {
            d1[i + 1]
        };

        if work0 + work1 < k || work0 + work1 == k && last_work + c >= next_work {
            println!("{}", i + 1);
        }
    }
}