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

int main(){
  int N,K,c;
  cin>>N>>K>>c;
  string S;
  cin>>S;
  vector<int> a(K),b(K);

  int i=0;
  int cnt=0;
  while(cnt<K && i<N){
    if(S[i]=='x') ++i;
    else{
      a[cnt]=i;
      ++cnt;
      i += c+1;
    }
  }
  i=N-1;
  cnt=K-1;
  while(cnt>=0 && i>=0){
    if(S[i]=='x') --i;
    else{
      b[cnt]=i;
      --cnt;
      i=i-(c+1);
    }
  }

  for(int i=0;i<K;++i){
    if(a[i]==b[i]) cout<<a[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;
  vector<int> a(K),b(K);

  int i=0;
  int cnt=0;
  while(cnt<K && i<N){
    if(S[i]=='x') ++i;
    else{
      a[cnt]=i;
      ++cnt;
      i += c+1;
    }
  }
  i=N-1;
  cnt=K-1;
  while(cnt>=0 && i>=0){
    if(S[i]=='x') --i;
    else{
      b[cnt]=i;
      --cnt;
      i=i-(c+1);
    }
  }

  for(int i=0;i<K;++i){
    if(a[i]==b[i]) cout<<a[i]+1<<endl;
  }

}
```