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 (1.15.1) (judge verdict: Time Limit Exceeded)

use std::collections::BTreeSet;
use std::io::{self, BufRead};

fn find_count(cal: &[u8], d: usize, k: usize, c: usize, memo: &mut Vec<Vec<Option<Box<BTreeSet<usize>>>>>) -> Option<Box<BTreeSet<usize>>> {
  if k == 0 {
    return Some(Box::new(BTreeSet::new()));
  }
  if cal.len() <= d {
    return None;
  }
  if let Some(cnt) = memo[d][k-1].clone() {
    return Some(cnt.clone());
  }
  const O: u8 = 'o' as u8;
  const X: u8 = 'x' as u8;
  let cnt = if cal[d] == O {
    let o1 = find_count(cal, d + c + 1, k - 1, c, memo);
    let o2 = find_count(cal, d + 1, k, c, memo);
    match (o1.clone(), o2.clone()) {
      (Some(s1), Some(s2)) => {
        Some(Box::new(s1.intersection(&s2).map(|x| x.clone()).collect::<BTreeSet<usize>>()))
      },
      (Some(s1), None    ) => {
        let mut s1_ : Box<BTreeSet<usize>> = Box::new((*s1).clone());
        s1_.insert(d);
        Some(s1_)
      },
      (None,     Some(s2)) => {
        let mut s2_ = Box::new((*s2).clone());
        s2_.insert(d);
        Some(s2_)
      }
      (None,     None    ) => None,
    }
  } else if cal[d] == X {
    find_count(cal, d + 1, k, c, memo)
  } else {
    panic!("unsupported char")
  };
  memo[d][k-1] = cnt.clone();
  cnt
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let l1 = lines.next().unwrap().unwrap();
  let v1: Vec<i32> = l1.split(' ').map(|x| x.parse().unwrap()).collect();
  let cal: String = lines.next().unwrap().unwrap();
  let n = v1[0] as usize;
  let k = v1[1] as usize;
  let c = v1[2] as usize;

  let mut memo: Vec<Vec<Option<Box<BTreeSet<usize>>>>> = vec![vec![None; k]; n];

  let cnt = find_count((&cal).as_bytes(), 0, k, c, &mut memo);

  if let Some(s) = cnt {
    for i in *s {
      println!("{}", i + 1);
    }
  }
}