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;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007,MOD2=998244353;



int main() {
    int N,K,C;
    string S;
    cin>>N>>K>>C>>S;
    vector<int> L,R,l(K),r(K);
    int point=0;
    while(point<N){
        
        if(S.at(point)=='o'){
            L.push_back(point);
            point+=C+1;
        }
        else if(S.at(point)=='x'){point++;}
    }
    rep(i,K){
        l.at(i)=L.at(i);
    }

    point=0;
    reverse(S.begin(),S.end());
    while(point<N){
        
        if(S.at(point)=='o'){
            R.push_back(point);
            point+=C+1;
        }
        else if(S.at(point)=='x'){point++;}
    }
    rep(i,K){
        r.at(i)=N-1-R.at(i);
    }
    reverse(r.begin(),r.end());
    int ans=0;
    rep(i,K){
        if(r.at(i)==l.at(i))cout<<r.at(i)+1<<endl;;
    }
}