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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program test
implicit none
integer(8) :: n,k,c,i,j,dp_left(0:200000)=10**6,dp_right(0:200000)=10**6,workday=0
character(len=200000) :: s

read(*,*) n,k,c
read(*,*) s(1:n)

!dp_leftは,左から見たときにi日目の最速の労働日
workday = 0
dp_left(0) = 0
dp_right(0) = 0

do i=1,n
  if(s(i:i)=='x')then
    continue
  else
    if(workday == 0)then
      workday = workday + 1
      dp_left(workday) = i
    else
      if(i > dp_left(workday) + c) then
        workday = workday + 1
        dp_left(workday) = i
      endif
    endif
  endif
enddo

!dp_rightは,右から見たときにi日目の最遅の労働日
workday = 0
do i=n,1,-1
  if(s(i:i)=='x')then
    continue
  else
    if(workday == 0)then
      workday = workday + 1
      dp_right(workday) = i
    else
      if(i < dp_right(workday) - c) then
        workday = workday + 1
        dp_right(workday) = i
      endif
    endif
  endif
enddo

do i=1,k
  if(dp_left(i)==dp_right(k+1-i)) then
    write(*,*) dp_left(i)
  endif
enddo

end