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 (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(java.util.Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun sum(N: Int, C: Int, S: BooleanArray): IntArray {
        val L = IntArray(N)
        var n = 0
        var i = 0
        while (i < N) {
            if (S[i]) {
                n++
                for (j in 0 until C) {
                    L[i] = n
                    i++
                    if (i >= N - 1) return L
                }
            }
            L[i] = n
            i++
        }
        return L
    }

    fun solve(sc: java.util.Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val K = sc.next().toInt()
        val C = sc.next().toInt()
        val S = sc.next().map { it == 'o' }

        val L = sum(N, C, S.toBooleanArray())
        val R = sum(N, C, S.reversed().toBooleanArray()).reversed().toIntArray()

        for (i in 0 until N) {
            val l = if (i > 0) L[i - 1] else 0
            val r = if (i + 1 < N) R[i + 1] else 0
            if (l + r < K) {
                pw.println(i + 1)
            }
        }
    }
}