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.chomp.split(" ").map(&:to_i)
@s = gets.chomp

def getPositions()
  res = []
  i = 0
  while i < @n do
    break if res.count >= @k
    if @s[i] == 'o'
      res.push(i)
      i += @c + 1
    else
      i += 1
    end
  end
  return res 
end

l = getPositions()
@s = @s.reverse
r = getPositions()
@k.times do |i|
  r[i] = @n - 1 - r[i]
end
@s = @s.reverse

lastL = Array.new(@n+1, -1)
@k.times {|i| lastL[l[i]+1] = i}
@n.times do |i|
  lastL[i+1] = lastL[i] if lastL[i+1] == -1
end

lastR = Array.new(@n+1, -1)
@k.times {|i| lastR[r[i]] = i}
(@n-1).downto(0) do |i|
  lastR[i] = lastR[i+1] if lastR[i] == -1
end

@n.times do |i|
  next if @s[i] == 'x'
  li = lastL[i]
  ri = lastR[i+1]
  cnt = 0
  cnt += li + 1 if li != -1
  cnt += ri + 1 if ri != -1
  cnt-=1 if li != -1 && ri != -1 && r[ri]-l[li] <= @c
  next if cnt >= @k
  puts i+1
end