#include <bits/stdc++.h>
using namespace std;

int main() {
  int N,K,C;
  cin>>N>>K>>C;
  string S;
  cin>>S;
  int a=0,b=0;
  vector<int>p(K);
  vector<int>q(K);
  while(a<K){
    if(S.at(b)=='x'){
      b++;
    }
    else{
      p.at(a)=b;
      b+=C+1;
      a++;
    }
  }
  
  a=K-1,b=N-1;
  while(a>=0){
    if(S.at(b)=='x'){
      b--;
    }
    else{
      q.at(a)=b;
      b-=C+1;
      a--;
    }
  }
  for(int i=0;i<K;i++){
    if(p.at(i)==q.at(i)){
      cout<<p.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>
using namespace std;

int main() {
  int N,K,C;
  cin>>N>>K>>C;
  string S;
  cin>>S;
  int a=0,b=0;
  vector<int>p(K);
  vector<int>q(K);
  while(a<K){
    if(S.at(b)=='x'){
      b++;
    }
    else{
      p.at(a)=b;
      b+=C+1;
      a++;
    }
  }
  
  a=K-1,b=N-1;
  while(a>=0){
    if(S.at(b)=='x'){
      b--;
    }
    else{
      q.at(a)=b;
      b-=C+1;
      a--;
    }
  }
  for(int i=0;i<K;i++){
    if(p.at(i)==q.at(i)){
      cout<<p.at(i)+1<<endl;
    }
  }
  
}
```