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 Ruby (2.3.3) (judge verdict: Accepted)

N, K, C = gets.split.map(&:to_i)
S = gets.strip.chars.map { |char| char == 'o' ? 0 : nil }
ss = S.size

# Justify the numbers to the right
dates_from_left = S
# Justify the numbers to the left
dates_from_right = S.reverse
ans = []

i = 0
plan = 1
while i < N
  break if plan > K
  if dates_from_left[i] == 0
    dates_from_left[i] = plan
    plan += 1
    i += C + 1
  else
    i += 1
  end
end

i = 0
plan = K
while  i < N
  break if plan < 1
  if dates_from_right[i] == 0
    dates_from_right[i] = plan
    plan -= 1
    i += C + 1
  else
    i += 1
  end
end

dates_from_left.each_with_index do |date, i|
  unless date.nil? || date == 0
    ans << i + 1 if date == dates_from_right[ss-i-1]
  end
end
if ans.nil?
  puts "\n"
else
  ans.each do |a|
    puts a
  end
end