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)

fn main() {
    let (n, k, c): (usize, usize, usize) = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
        )
    };

    let s: Vec<char> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().chars().collect()
    };

    let mut left = vec![0; k];
    let (mut li, mut lj) = (0, 0);
    while li < n && lj < k {
        if s[li] == 'o' {
            left[lj] = li;
            li += c;
            lj += 1;
        }
        li += 1;
    }

    let mut right = vec![0; k];
    let (mut ri, mut rj) = (0, 0);
    while ri < n && rj < k {
        if s[n - 1 - ri] == 'o' {
            right[k - 1 - rj] = n - 1 - ri;
            ri += c;
            rj += 1;
        }
        ri += 1;
    }

    for i in 0..k {
        if left[i] == right[i] {
            println!("{}", left[i] + 1);
        }
    }
}