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 Python (3.8.2) (judge verdict: Wrong Answer)

from collections import deque
import copy

N, K, C = map(int,input().split())
S = input().split()[0]
TW = 0
q = []

for s in range(len(S)):
    if S[s] == 'o':
        TW += 1
        q.append(s+1)

if TW == K:
    for i in range(len(q)):
        print(q[i])

WD = []
LD = 0

for s in q:
    if len(WD) > K:
        break
    if LD == 0:
        WD.append(s)
        LD = s
    elif s-LD <= C:
        continue
    else:
        WD.append(s)
        LD = s


WDR = []
LD = 0
q.reverse()

for s in q:
    if len(WDR) > K:
        break
    if LD == 0:
        WDR.append(s)
        LD = s
    elif LD - s <= C:
        continue
    else:
        WDR.append(s)
        LD = s


for d in range(len(WD)):
    if WD[d] == WDR[len(WD)-d-1]:
        print(WD[d])