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.chomp.split("")

wd = Array.new(n).map{Array.new(3,0)}

#top
cnt = 0
pre = c
for i in 0..(n-1)
  pre += 1
  if s[i]=="o" && pre > c
    wd[i][0] = 1
    pre = 0
    cnt += 1
    break if cnt >= k
  else
    wd[i][0] = 0
  end
end

#bottom
cnt = 0
pre = c
id = n-1
while id >= 0
  pre += 1
  if s[id]=="o" && pre > c
    wd[id][1] = 1
    pre = 0
    cnt += 1
    break if cnt >= k
  else
    wd[id][1] = 0
  end
  id -= 1
end

#both
cnt = 0
tpre = c
bpre = c
tid = 0
bid = n-1
cmp = -1
for i in 1..((k+1)/2)
  while 1
    tpre += 1
    if s[tid]=="o" && tpre > c
      wd[tid][2] = 1
      tpre = 0
      cnt += 1
      cmp = tid if cnt >= k
      tid += 1
      break
    else
      wd[tid][2] = 0
    end
    tid += 1
  end
  while 1
    bpre += 1
    if s[bid]=="o" && bpre > c
      wd[bid][2] = 1
      bpre = 0
      cnt += 1
      if cnt >= k && cmp > -1
        if cmp != bid
          wd[cmp][2] = 0
          wd[bid][2] = 0
        end
      end
      bid -= 1
      break
    else
      wd[bid][2] = 0
    end
    bid -= 1
  end
end

for i in 0..(n-1)
  puts i+1 if wd[i][0]*wd[i][1]*wd[i][2] == 1
end