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)

#require 'prime'
#require 'openssl'
MOD=10**9+7;  cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.chomp.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def ar2(s1,s2,de=nil) Array.new(s1){Array.new(s2,de)} end
def ar3(s1,s2,s3,de=nil) Array.new(s1){Array.new(s2){Array.new(s3,de)}} end

def desc(ar) ar.sort!{|x,y| y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+).to_i end
def ceil(a,b) (a.to_f/b).ceil end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end
def rui(ar) s=[0]; ar.size.times{|i| s<<s[-1]+ar[i]}; s end
def rev_rui(ar) s=[0]; ar.size.times.reverse_each{|i| s<<s[-1]+ar[i]}; s end

def puts_yesno(b) puts(b ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def pu(h) begin; require './../nemi_lib/debug.rb'; puts_val(h); rescue LoadError; end; end
########### ( ˘ω˘ )ｽﾔｧ… ###########
n,k,c=gsmi
s=gs

#i日目までに働ける最大の数
rui=[]
idx=-10**9
n.times do |i|
  tar=s[i]
  if i<=idx+c
    # 間空いてない
    rui<< rui[-1].to_i
  else
    if tar=="o"
      rui<< rui[-1].to_i + 1
      idx=i
    else
      rui<< rui[-1].to_i
    end
  end
end

# 右から
r_rui=[]
idx=-10**9
sr=s.reverse
n.times do |i|
  tar=sr[i]
  if i<=idx+c
    # 間空いてない
    r_rui<< r_rui[-1].to_i
  else
    if tar=="o"
      r_rui<< r_rui[-1].to_i + 1
      idx=i
    else
      r_rui<< r_rui[-1].to_i
    end
  end
end

pu rui:rui.to_s
pu r_rui:r_rui.reverse.to_s

ar=rui.zip(r_rui.reverse).map{|i|sum i}
pu ar:ar.to_s

# 左からと右からの２つを足して、k+1になった前後でkになってればいい。
ans=[]
prev=-1
n.times do |i|
  if ar[i]>=k+1 && prev<=k && ar[i+1].to_i<=k
    ans<<i+1
  end
  prev=ar[i]
end

puts ans