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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
$input = explode(' ', trim(fgets(STDIN)));

$n = (int)$input[0];
$k = (int)$input[1];
$c = (int)$input[2];
$s = trim(fgets(STDIN));
// 勤務可能日の累積和
// $o_sum = [0];
// for ($i=0; $i < $n; $i++) { 
// 	$prev = $o_sum[$i];
// 	if ($s[$i] == "o") {
// 		$o_sum[] = $prev + 1;
// 	} else {
// 		$o_sum[] = $prev;
// 	}
// }

// 予定を前から詰めていった場合と、後から詰めていった場合とで、一致する日を確認する（それは必ず働く日である可能性がある）
$from_front = [];
$cur = 0;
while (count($from_front) < $k) {
	if ($s[$cur] == "o") {
		$from_front[] = ($cur + 1);
		$cur += $c + 1;
	} else {
		$cur += 1;
	}
}
$from_back = [];
$cur = $n - 1;
while (count($from_back) < $k) {
	if ($s[$cur] == "o") {
		$from_back[] = ($cur + 1);
		$cur -= ($c + 1);
	} else {
		$cur -= 1;
	}
}
sort($from_back);

$result = [];
foreach ($from_front as $idx => $fr) {
	if ($from_back[$idx] == $fr) {
		$result[] = $fr;
	}
}
echo implode(PHP_EOL, $result);