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
// E 解説見て
fscanf(STDIN, '%d %d %d', $n, $k, $c);
fscanf(STDIN, '%s', $s);

// 左から
$rest = 10 ** 6; // 今日まで連続で休んだ日数の初期値は十分大きい値
$cnt = 1; // 何回目に働くか
for ($i = 1; $i <= $n && $cnt <= $k; $i++) {
    if ($s[$i - 1] === 'o' && $rest >= $c) {
        // $cnt 回目に働くのは $i 日目以降
        $l[$cnt] = $i;
        $cnt++;
        $rest = 0;
    } else {
        $rest++;
    }
}

// 右から
$rest = 10 ** 6; // 今日まで連続で休んだ日数の初期値は十分大きい値
$cnt = $k; // 何回目に働くか
for ($i = $n; $i >= 1 && $cnt >= 1; $i--) {
    if ($s[$i - 1] === 'o' && $rest >= $c) {
        // $cnt 回目に働くのは $i 日目以前
        $r[$cnt] = $i;
        $cnt--;
        $rest = 0;
    } else {
        $rest++;
    }
}

$ans = [];
for ($i = 1; $i <= $k; $i++) {
    if (isset($l[$i]) && isset($r[$i]) && $l[$i] === $r[$i]) $ans[] = $l[$i];
}
echo implode(PHP_EOL, $ans);