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::{
   fastout, input,
   marker::{Bytes, Chars, Isize1, Usize1},
};

#[fastout]
fn main() {
   input! {
      n:usize,k:usize,c:usize,
      s:Chars
   }

   let mut s1 = vec![];
   let mut s2 = vec![];
   let mut interval = c + 1;
   let mut count = 0;

   for (i, x) in s.iter().enumerate() {
      if *x == 'o' && interval > c && count < k {
         interval = 0;
         s1.push(i + 1);
         count += 1;
      }
      interval += 1;
   }

   count = 0;
   interval = c + 1;

   for (i, x) in s.iter().enumerate().rev() {
      if *x == 'o' && interval > c && count < k {
         interval = 0;
         s2.push(i + 1);
         count += 1;
      }
      interval += 1;
   }

   s2.reverse();
   //dbg!(&s1, &s2);
   for i in 0..k {
      if s1[i] == s2[i] {
         println!("{}", s1[i]);
      }
   }
}