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

int main() {

  int N, K, C;
  cin >> N >> K >> C;

  string S;
  cin >> S;

  vector<int> saitan(K);
  vector<int> saichi(K);

  int j = 0;
  for(int i = 0; i < N && j < K; i++){
    if(S[i] == 'o'){
      saitan[j] = i;
      i += C;
      j++;
    }
  }

  j = 0;
  for(int i = 0; i < N && j < K; i++){
    if(S[N - 1 - i] == 'o'){
      saichi[K - 1 - j] = N - 1 - i;
      i += C;
      j++;
    }
  }

  for(int i = 0; i < K; i++){
    if(saitan[i] == saichi[i]){
      cout << saitan[i] + 1 << endl;
    }
  }
}
### Prompt

Please create a solution in CPP 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;
  cin >> N >> K >> C;

  string S;
  cin >> S;

  vector<int> saitan(K);
  vector<int> saichi(K);

  int j = 0;
  for(int i = 0; i < N && j < K; i++){
    if(S[i] == 'o'){
      saitan[j] = i;
      i += C;
      j++;
    }
  }

  j = 0;
  for(int i = 0; i < N && j < K; i++){
    if(S[N - 1 - i] == 'o'){
      saichi[K - 1 - j] = N - 1 - i;
      i += C;
      j++;
    }
  }

  for(int i = 0; i < K; i++){
    if(saitan[i] == saichi[i]){
      cout << saitan[i] + 1 << endl;
    }
  }
}
```