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)

n,k,c = map(int,input().split())
s = input()

if c == 0:
    ls = [i for i in range(n) if s[i]=='o']
    if len(ls) == k:
        for i in ls:
            print(i+1)
    else:
        print()
    exit(0)

dp = [[0]*2 for i in range(n+1)]  # dp[i][0/1] i日目が(x/o)のときの最大値

for i in range(n-1,-1,-1):
    dp[i][0] = max(dp[i+1][0], dp[i+1][1])
    if s[i] == 'o':
        if i+c+1 >= n:
            dp[i][1] = 1
        else:
            dp[i][1] = max(dp[i+c+1][0], dp[i+c+1][1]) + 1

if max(dp[0][0],dp[0][1]) == k:
    ls = [[] for i in range(k+2)]
    for i in range(n):
        x = dp[i][1]
        if x >= 1:
            if len(ls[x+1]) == 1 and i+1 <= ls[x+1][0]+c:
                continue
            ls[dp[i][1]].append(i+1)
    for l in ls:
        if len(l) == 1:
            print(l[0])