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: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
int N,K,C;
int A[200001];
int B[200001];
vector<int> ans;



int main(){
    int k = 0;
    int cntt=0;
    cin >> N >> K >> C;
    int cnt=10000001;
    for(int i=0;i<N;i++){
        char s;
        cin >> s;
        if (s == 'o' && cnt < C){
            A[i] = 1;
            cnt++;
        } else if (s == 'x'){
            A[i] = 0;
            cnt++;
        } else {
            A[i] = 2;
            B[k] = cnt;
            cnt = 0;
            k++;
            cntt = i;
        }
    }
    B[k] = N-cntt-1;
    
    if (k>K){
        return 0;
    }
    int i=N-1;
    int cnttt = C;
    while(k>=0){
        for (int j=0;j<B[k]+1;j++){
            if(A[i-j] == 1){
                if(C-j <= cnttt){
                    cnttt = B[k]-j;
                    break;
                }
            }else if(A[i-j] == 2){
                cnttt = 0;
                ans.push_back(i-j+1);
            }
        }
        i -= (B[k]+1);
        k--;
    }
    for(ll j=ans.size()-1;j>=0;j--){
        cout << ans[j] << endl;
    }
    
}