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 <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <vector>

using namespace std;

#define ll long long
#define db double
#define clr(a, b) memset(a, b, sizeof(a))

const int maxn = 5e5 + 10;
const int inf = 0x7ffffff;

int n, k, c;
void solve()
{
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<int> vec;
    set<int> rvec;
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        if (s[i] == 'o')
        {
            vec.push_back(i);
            i += c;
        }
        if (vec.size() == k+1)
            return ;
    }
    for (int i = n - 1; i >= 0; i--)
    {
        if (s[i] == 'o')
        {
            rvec.insert(i);
            i -= c;
            cnt++;
        }
        if (rvec.size() == k+1)
            return ;
    }
    for (int i:vec)
    {
        int j = i;
        if(rvec.count(j))
        {
            cout << i + 1 << "\n";
        }
    }
}

int main()
{
    //freopen("out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    //cin >> T;
    while (T-- > 0)
        solve();
}