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: Accepted)

import sys

sys.setrecursionlimit(10 ** 7)
rl = sys.stdin.readline


def solve():
    N, K, C = map(int, rl().split())
    S = input()
    
    left_greedy = [False] * N
    right_greedy = [False] * N
    
    i, cnt = 0, 0
    while i < N:
        if S[i] == 'o':
            cnt += 1
            left_greedy[i] = True
            i += C + 1
        else:
            i += 1
    
    if K < cnt:
        exit()
    
    i = N - 1
    while 0 <= i:
        if S[i] == 'o':
            right_greedy[i] = True
            i -= C + 1
        else:
            i -= 1
    
    for i in range(N):
        if left_greedy[i] and right_greedy[i]:
            print(i + 1)


if __name__ == '__main__':
    solve()