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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!(
    n: usize,
    k: usize,
    c: usize,
    s: chars,
    );
    let mut l = vec![];
    let mut r = vec![];
    let mut h = s.iter().map(|&x| x == 'o').collect::<Vec<bool>>();
    let mut i = 0;
    let mut t = 0;
    while t < k {
        if h[i] {
            l.push(i);
            t += 1;
            i += c + 1;
        } else {
            i += 1;
        }
    }
    h.reverse();
    i = 0;
    t = 0;
    while t < k {
        if h[i] {
            r.push(i);
            i += c + 1;
            t += 1;
        } else {
            i += 1;
        }
    }
    r.reverse();
    for (&i, &j) in l.iter().zip(r.iter()) {
        if i == n - 1 - j {
            println!("{}", i + 1);
        }
    }
}