#include <bits/stdc++.h>
#define rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll MOD = 1000000007;

    

int main(){
  ll N,K,C,t=0;
  string S;
  cin >> N >> K >> C >> S;
  vector<ll> L(K),R(K);
  rep(i,N){
    if(S.at(i)=='o'){
      L.at(t)=i;
      t++;
      i+=C;
    }
    if(t==K){
      break;
    }
  }
  t=K-1;
  rep(i,N){
    if(S.at(N-i-1)=='o'){
      R.at(t)=N-i-1;
      t--;
      i+=C;
    }
    if(t==-1){
      break;
    }
  }
  rep(i,K){
    if(L.at(i)==R.at(i)){
      cout << L.at(i)+1 << endl;
    }
  }
      
}
### Prompt

Construct a cpp code solution to the problem outlined:

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.

Examples

Input

11 3 2
ooxxxoxxxoo


Output

6


Input

5 2 3
ooxoo


Output

1
5


Input

5 1 0
ooooo


Output




Input

16 4 3
ooxxoxoxxxoxoxxo


Output

11
16

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll MOD = 1000000007;

    

int main(){
  ll N,K,C,t=0;
  string S;
  cin >> N >> K >> C >> S;
  vector<ll> L(K),R(K);
  rep(i,N){
    if(S.at(i)=='o'){
      L.at(t)=i;
      t++;
      i+=C;
    }
    if(t==K){
      break;
    }
  }
  t=K-1;
  rep(i,N){
    if(S.at(N-i-1)=='o'){
      R.at(t)=N-i-1;
      t--;
      i+=C;
    }
    if(t==-1){
      break;
    }
  }
  rep(i,K){
    if(L.at(i)==R.at(i)){
      cout << L.at(i)+1 << endl;
    }
  }
      
}
```