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: Wrong Answer)

n,k,c=gets.split(' ').map(&:to_i)
#h=gets.split(' ').map(&:to_i)
#tajigen = Array.new(4) { Array.new(2,0) }
s=gets.chomp

fast = Array.new(n+1,0)
latest = Array.new(n+1,0)

u=0
for  i in 0..k-1
    for j in u..n-1
      if s[j]=='o'
        fast[j+1] = 1
        u = j+c+1
        break
      end
    end
end

u=0
s.reverse!
for i in 0..k-1
  #for j in (0..u).to_a.reverse #for分で降順
  for j in u..n-1
    if s[j] == 'o'
      latest[j] = 1
      u = j+c+1
      break
    end
  end
end
latest.reverse!
#puts latest
#latest.reverse!
ans=Array.new
=begin
n=0
while n<fast.length
  if latest.include?(fast[n])
    ans.push(fast[n])
  end
  n+=1
end
=end
=begin
p=0
q=0
while p<k && q<k
  if fast[p]==latest[q]
    ans.push(fast[p])
    p+=1
    q+=1
  else
    p+=1
  end
end
=end
#puts fast
for i in 1..n
  if fast[i]==1 && latest[i]==1
    ans << i
  end
end
puts ans