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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int n, k, c;
string s;
vector<int> l, r;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    
    cin >> n >> k >> c;
    cin >> s;
    int cnt = 0;
    for (int i = 0; i < n && cnt < k; i++)
        if (s[i] == 'o' && (l.empty() || l.back() + c < i)){
            l.push_back(i);
            cnt++;
        }
    
    cnt = 0;
    for (int i = n - 1; i >= 0 && cnt < k; i--)
        if (s[i] == 'o' && (r.empty() || r.back() - c > i)){
            r.push_back(i);
            cnt++;
        }

    for (int i = 0; i < k; i++)
        if (l[i] == r[k - i - 1])
            cout << l[i] + 1 << "\n";
    
    return 0;
}