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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

n,k,c = map(int,input().split())
s = list(input())
left = [0]*n
right = [0]*n
s_r = list(reversed(s))

count_l=1
count_r=1
i=j=0
while i<n or j<n:
    
    if i >= n:
        i=n
    else:
        if s[i] == 'o':
            left[i] = count_l
            count_l+=1
            i+=c+1
        else:
            i+=1
            
    if j >= n:
        j=n
    else:    
        if s_r[j] == 'o':
            right[j] = count_r
            count_r += 1
            j+=c+1
        else:
            j+=1
    
right.reverse()
need=[]
for i in range(n):
    if s[i] != 'o':
        continue
    else:
        j=h=0
        cl=cr=0
        if i!=0:
            while i-j>=0:
                j+=1
                if left[i-j] != 0:
                    cl = left[i-j]
                    cl_i = i-j
                    break
        else:
            cl_i=-c
        
        if i!=n-1:
            while i+h <= n-1:
                h+=1
                if right[i+h] != 0:
                    cr = right[i+h]
                    cr_i = i+h
                    break
        else:
            cr_i=n+c
    if cr_i-cl_i<=c:
        co = cr+cl-1
        
    else:
        co = cr+cl

    if co<k:
        need.append(str(i+1))


print('\n'.join(need))