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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define _GLIBCXX_DEBUG
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
template<class T> istream& operator >> (istream& s, vector<T>& v)
{ for (T& x: v) { s >> x; } return s;}

int n,k,c;
string s;
map<int, vector<int>> able;
vector<int> memo, memo2;
map<int,int> m;
int all = 0;

void check(int date, int pre, vector<int> list){
    list.push_back(date);
    if(pre == k){
        for(int i : list){
            if(m[i] == 0) memo2.push_back(i);
            m[i]++;
        }
        all++;
    }
    for(int i: able[date]) check(i ,pre+1, list);
}

int main(){
    cin >> n >> k >> c >> s;
    int size = (int)s.size();
    rep(i,size){
        if(s[i] == 'o'){
            memo.push_back(i);
            for(int j = i+c+1; j < size; j++) if(s[j] == 'o') able[i].push_back(j);
        }
    }
    vector<int> none;
    for(int i: memo) if(able[i].size() >= k-1) check(i, 1 , none);
    sort(memo2.begin(), memo2.end());
    for(int i: memo2) if(all == m[i]) cout << i+1 << endl;
    return 0;
}