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: Time Limit Exceeded)

# coding: utf-8

def greedy_search(K, C, workable, shift = 1):
    workable = workable[::shift]
    if workable == []:return []
    work_day = 1
    pre_day = workable[0]
    work_list = [workable[0]]
    if work_day == K:
        return work_list
    if workable[1:] == []:
        return []
    for work in workable[1:]:
        if shift*(pre_day + C *shift) >= work*shift:
            continue
        pre_day = work
        work_day += 1
        work_list.append(work)
        if work_day == K:
            return work_list
    return []

def extract(S):
    ret = []
    for idx, s in enumerate(S, 1):
        if s == 'o':
            ret.append(idx)
    return ret

if __name__ == '__main__':
    N, K, C = list(map(int, input().split()))
    S = list(input())
    base = extract(S)
    left = greedy_search(K,C,base,shift=1)
    right= greedy_search(K,C,base,shift=-1)[::-1]
    #print(base)
    #print(left)
    #print(right)
    for i in base:
        left_k = len(left)
        left_margin = left[-1] if left is not [] else -10**10
        for left_idx in range(len(left)):
            if left[left_idx] >= i:
                left_k = left_idx
                left_margin = left[left_idx-1] if left_idx != 0 else -10**10
                break
        right_k = 0
        right_margin = 10 ** 10
        for right_idx in range(len(right)):
            if right[right_idx] > i:
                right_k = len(right) - right_idx
                right_margin = right[right_idx]
                break
        kk = left_k + right_k
        if right_margin - left_margin <= C:
            kk -= 1
        if kk < K:
            print(i)