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 Python (3.4.3) (judge verdict: Runtime Error)

N, K, skip = map(int, input().split(" "))
lazy = list(input())
skip_to = [None]*N
for i in range(N)[::-1] : 
    if lazy[i] == "o" :
        skip_to[i] = i
    else :
        if i == N-1 :
            continue
        skip_to[i] = skip_to[i+1]

must_work = [True]*N

def check (work_day, i, N, K, skip) :
    if not any(must_work) :
        return True
    if len(work_day) == K :
        m = 0
        for j in range(N) : 
            if m < K and j ==  work_day[m] : 
                must_work[j] = must_work[j] and True
                m += 1
            else :
                must_work[j] = False
        return False

    if i >= N :
        return False

    if lazy[i] == "x" :
        check(work_day, skip_to[i], N, K, skip)
    else :
        work_day.append(i)
        if check(work_day, i+skip, N, K, skip) :
            return True
        work_day.pop()
        
    if check(work_day, i+1, N, K, skip) :
        return True
    return False

for i in range(N) : 
    if check([], i, N, K, skip) :
        break 

for i in range(N) :
    if must_work[i] :
        print(i+1)