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: Accepted)

N,K,C = map(int,input().split())
S = input()
def solve():
    if C==0:
        a = [i+1 for i,c in enumerate(S) if c=='o']
        if len(a)==K:
            print(*a, sep='\n')
        else:
            print()
        exit()
    l = []
    y = 0
    for i,c in enumerate(S):
        if y:
            y -= 1
            continue
        if c=='o':
            y = C
            l.append(i+1)
            if len(l) == K:
                break
    r = []
    y = 0
    for i,c in enumerate(S[::-1]):
        if y:
            y -= 1
            continue
        if c=='o':
            y = C
            r.append(N-i)
            if len(r) == K:
                break
    r.reverse()
    ans = [a for a,b in zip(l,r) if a==b]
    print(*ans, sep='\n')
solve()