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 copy

n,k,c = map(int, input().split())
s = input()

ls = []
for i in range(n) :
    if s[i] == 'o' :
        ls.append([i])
subLs = []
while(len(ls) > 0) :
    newLs = ls.pop()
    if len(newLs) >= k :
        subLs.append(newLs)
    else :
        last = newLs.pop()
        canTo = []
        for i in range(last+c,n) :
            if s[i] == 'o' :
                canTo.append (i)
        while len(canTo) > 0 :
            tmpLs = newLs.copy()
            tmpLs.append(last)
            tmpLs.append(canTo.pop())
            ls.append(tmpLs)
checkLs = sum(subLs,[])
dupLs = list(set(checkLs))
mx = 0
ans = 0
for i in dupLs :
    cnt = checkLs.count(i)
    if mx < cnt :
        mx = cnt
        ans = i
print(ans+1)