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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<stdio.h>
#include<string>
#include<vector>
#include <cmath>
#define rep(i,n) for (int i=0;i<n;i++)
//#include <bits/stdc++.h>
using namespace std;
//void narabekae(string s[]){
    
//}


int main(){
    int N,K,C;
    cin>>N>>K>>C;
    string S;
    cin>>S;
    //for (int i=0;i<N;i++) cin>>S[i];
    vector<int> v;
    int j=0,k=0;
    while(k<K&&j<N){
        if (S[j]=='o'){
            //cout<<j<<" ";
            v.push_back(j);
            j+=C+1;
            k+=1;
        }
        else j+=1;
    }
    vector<int> w;
    j=N-1;
    k=0;
    while(k<K&&j>=0){
        if (S[j]=='o'){
            //cout<<j<<" ";
            w.push_back(j);
            j-=C+1;
            k+=1;
        }
        else j-=1;
    }
    rep(i,K){
        if (v[i]==w[K-1-i]) cout<<v[i]+1<<endl;
    }
    return 0;
}