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

int main() {
  int N,K,C;
  string S;
  cin >> N >> K >> C >> S;
  vector<int> L(K);
  vector<int> B(K);
  
  int cnt = 0;
  for(int i=0;i < N;i++){
    if(S.at(i) == 'o'){
      L.at(cnt) = i;
      cnt++;
      i += C;
    }
    if(cnt >= K){
      break;
    }
  }
  cnt = K - 1;
  for(int i=0;i < N;i++){
    if(S.at(N - i - 1) == 'o'){
       B.at(cnt) = N - i - 1;
       cnt--;
       i +=C;
    }
    if(cnt < 0){
      break;
    }
  }
  for(int i=0;i < K;i++){
    if(L.at(i) == B.at(i)){
      cout << L.at(i) + 1 << endl;
    }
  }
}
### Prompt

Please formulate a cpp solution to the following problem:

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;
  string S;
  cin >> N >> K >> C >> S;
  vector<int> L(K);
  vector<int> B(K);
  
  int cnt = 0;
  for(int i=0;i < N;i++){
    if(S.at(i) == 'o'){
      L.at(cnt) = i;
      cnt++;
      i += C;
    }
    if(cnt >= K){
      break;
    }
  }
  cnt = K - 1;
  for(int i=0;i < N;i++){
    if(S.at(N - i - 1) == 'o'){
       B.at(cnt) = N - i - 1;
       cnt--;
       i +=C;
    }
    if(cnt < 0){
      break;
    }
  }
  for(int i=0;i < K;i++){
    if(L.at(i) == B.at(i)){
      cout << L.at(i) + 1 << endl;
    }
  }
}
```