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)


import sys, io, os
input = sys.stdin.readline
inputt = lambda: range(int(input()))
inputn = lambda n = 1: range(n)
inputi = lambda: map(int, input().split())
inputl = lambda: list(inputi())
from math import *
from heapq import *
from bisect import *
from functools import reduce, lru_cache
from collections import Counter, defaultdict
from builtins import *



N, K, C = inputi()
S = list(*input().split())
#print(S)
S = [False] + [c == "o" for c in S]

@lru_cache(None)
def F(d):
    if d > N:
        return 0
    if not S[d]:
        return F(d + 1)
    return 1 + F(d + C + 1)


if F(0) >= K:
    i = 1
    while i <= N:
        #print(i, N, S)
        if S[i]:
            #print(i, K, F(i + 1))
            if F(i + 1) == K - 1:
                print(i)
            K -= 1
            i += C
        i += 1