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 (3.4.3) (judge verdict: Wrong Answer)

a,b,c = map(int,(input().split()))
mb = input()

rl = list()
ll = list()

for i in range(a):
    if mb[i] == "x":continue
    if len(rl) == 0:
        rl.append(i + 1)
    else:
        if rl[-1] + c + 1 <= i + 1:
            rl.append(i + 1)
            if len(rl) == b:break

for i in range(a-1,-1,-1):
    if mb[i] == "x":continue
    if len(ll) == 0:
        ll.append(i + 1)
    else:
        d1 = ll[-1] - c - 1
        d2 = i + 1
        if  d1 >= d2:
            ll.append(i + 1)
            if len(ll) == b:break

ans=list()
for i,k in zip(rl,reversed(ll)):
    if b == 1 or b == 0:break
    if i == k:
        ans.append(i)

[print(i) for i in ans]