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: Runtime Error)

fun main(args: Array<String>) {
    var (n,k,c)=readLine()!!.split(' ').map{it.toInt()}
    var s= readLine()!!
    var forward=Array(n){0}.toMutableList()
    var back=Array(n){0}.toMutableList()
    var tmp_f=c
    var tmp_b=c
    var ans=""
    var ind_f=0
    var ind_b=n-1
    for(i in 0 until n){
        if(tmp_f<c||s[i]=='x'){
            tmp_f++
        }else{
            tmp_f=0
            forward[ind_f]=i+1
            ind_f++
        }
        if(tmp_b<c||s[n-1-i]=='x'){
            tmp_b++
        }else{
            tmp_b=0
            back[ind_b]=n-i
            ind_b--
        }
    }
    for(i in 0 until n){
        if (forward[i]==0){
            break
        }
        if(forward[i]==back[n-1-i]){
            ans+="\n"+forward[i].toString()
        }
    }
    if(ans.length/2>k) println() else println(ans.substring(1))
}