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: Accepted)

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

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let ds: Vec<i32> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let s: Vec<char> = lines.next().unwrap().unwrap().chars().collect();
  let n = ds[0];
  let k = ds[1];
  let c = ds[2];

  let mut l = vec![-1 as i32; k as usize];
  let mut d: i32 = 0;
  let mut w: i32 = 0;
  while d < n && w < k {
    if s[d as usize] == 'o' {
      l[w as usize] = d;
      w += 1;
      d += c + 1;
    } else {
      d += 1;
    }
  }

  let mut r = vec![-1 as i32; k as usize];
  let mut d: i32 = n - 1;
  let mut w: i32 = k - 1;
  while d >= 0 && w >= 0 {
    if s[d as usize] == 'o' {
      r[w as usize] = d;
      w -= 1;
      d -= c + 1;
    } else {
      d -= 1;
    }
  }

  for i in 0..k {
    if l[i as usize] == r[i as usize] {
      println!("{}", l[i as usize] + 1);
    }
  }
}