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)

k = int(input())
lunluns = list(map(str, range(1,10)))
new = list(map(str, range(1,10)))
new_fueta = 9
end = False
while len(lunluns) < k:
  new_new_fueta = 0
  for x in lunluns[-new_fueta:]:
    str_x = str(x)
    if str_x[-1] == '0':
      lunluns += [int(str_x+'0'),int(str_x+'1')]
      new_new_fueta += 2
    elif str_x[-1] == '9':
      lunluns += [int(str_x+'8'),int(str_x+'9')]
      new_new_fueta += 2
    else:
      l = int(str_x[-1])
      lunluns += [int(str_x+str(l-1)), int(str_x+str(l)), int(str_x+str(l+1))]
      new_new_fueta += 3
  if end:
    break
  new_fueta = new_new_fueta
print(lunluns[k-1])