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

int main(){
  int n,k,c;
  cin >> n >> k >> c;
  string str;
  cin >> str;
  
  //cout << n << endl;
  int L[k], R[k];

  int p=0, num=0;

  while(num<k){
    if(str[p]=='o'){
      L[num] = p;
      //cout << p << endl;
      p+=c;
      num++;
      
    }
    p++;
  }
  
  num = k-1;
  p=n-1;
  while(num>=0){
    if(str[p]=='o'){
      R[num] = p;
      //cout << p << endl;
      p-=c;
      num--;
    }
    p--;
  }

  for(int i=0; i<k; i++){
    //cout << L[i] << " " << R[i] << endl;
    if(L[i]==R[i]){
      cout << L[i]+1 << endl;
    }
  }
  return 0;
}
### 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;
  cin >> n >> k >> c;
  string str;
  cin >> str;
  
  //cout << n << endl;
  int L[k], R[k];

  int p=0, num=0;

  while(num<k){
    if(str[p]=='o'){
      L[num] = p;
      //cout << p << endl;
      p+=c;
      num++;
      
    }
    p++;
  }
  
  num = k-1;
  p=n-1;
  while(num>=0){
    if(str[p]=='o'){
      R[num] = p;
      //cout << p << endl;
      p-=c;
      num--;
    }
    p--;
  }

  for(int i=0; i<k; i++){
    //cout << L[i] << " " << R[i] << endl;
    if(L[i]==R[i]){
      cout << L[i]+1 << endl;
    }
  }
  return 0;
}
```