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 Kotlin (1.3.71) (judge verdict: Accepted)

fun main() {
    val builder = StringBuilder()

    val min = -3_000_000

    val (n, k, c) = readInputLine().split(" ").map { it.toInt() }

    val s = readInputLine().toCharArray()

    var prev = min
    var cnt = 0

    val front = IntArray(k)

    for ((i, work) in s.withIndex()) {
        if (work == 'o' && i - prev > c) {
            prev = i
            front[cnt] = i + 1
            cnt++
            if (cnt == k) {
                break
            }
        }
    }

    prev = min
    cnt = 0

    val back = IntArray(k)

    for ((i, work) in s.reversed().withIndex()) {
        if (work == 'o' && i - prev > c) {
            prev = i
            back[k - 1 - cnt] = n - i
            cnt++
            if (cnt == k) {
                break
            }
        }
    }

    for (i in 0 until k) {
        if (front[i] == back[i]) {
            builder.appendln(front[i])
        }
    }

    print(builder.toString())
}

fun readInputLine(): String {
    return readLine()!!
}