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: Wrong Answer)

# import time

N, K, C = list(map(int, input().split()))
S = list(input())

F = []
L = []

# t1 = time.time()
# for i in range(N):
#     if S[i] == 'o':
#         if len(F) == 0:
#             F.append(i)
#         elif F[-1] + C < i:
#             F.append(i)
#     if len(F) >= K:
#         break
# print(F)
# t2 = time.time()
i = 0
while len(F) < K and i < N:
    if S[i] == 'o':
        if len(F) == 0:
            F.append(i)
        elif F[-1] + C < i:
            F.append(i)
    i += 1
# print(F)
# t3 = time.time()

# print(t2 - t1, t3 - t2)

i = N - 1
while len(L) < K and i >= 0:
    if S[i] == 'o':
        if len(L) == 0:
            L.append(i)
        elif L[-1] - C > i:
            L.append(i)
    i -= 1

ans = [i + 1 for i in set(F) & set(L)]
if len(ans) != 0:
    ans.sort()
    [print(i) for i in ans]
else:
    print()