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::Read;
use std::io::Write;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let k: usize = it.next().unwrap().parse().unwrap();
    let c: usize = it.next().unwrap().parse().unwrap();
    let s: Vec<char> = it.next().unwrap().chars().collect();
    let mut dp = vec![0; n + c + 1];
    for (i, x) in s.iter().enumerate().rev() {
        dp[i] = dp[i + 1];
        if *x == 'o' {
            dp[i] = std::cmp::max(dp[i], dp[i + 1 + c] + 1);
        }
    }
    if dp[0] > k {
        return;
    }
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    let mut i = 0;
    while i < n {
        while i < n && s[i] == 'x' {
            i += 1;
        }
        if dp[i] > dp[i + 1] {
            writeln!(out, "{}", i + 1).ok();
        }
        i += c + 1;
    }
}

fn main() {
    run();
}