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 (2.4.0) (judge verdict: Wrong Answer)

N, K, C = map(int, input().split())
S = input()

dp = [0] * N

dp[N - 1] = 1 if S[N - 1] == "o" else 0
pm = dp[N - 1]
for i in range(N - 2, -1, -1):
    if S[i] == "x" or N - i <= C + 1:
        dp[i] = dp[i + 1]
    else:
        dp[i] = dp[i + C + 1] + 1

# print(dp)
k = K
i = 0
cnt = 0
ans = 0
while i < N:
    if dp[i] < k:
        if cnt == 1:
            print(ans + 1)
            i = ans + C + 1
        cnt = 0
        k -= 1
        if k == 0:
            break
    if S[i] == "o" and dp[i] >= k:
        cnt += 1
        if cnt == 1:
            ans = i
    i += 1

if cnt == 1:
    print(ans + 1)