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

int main() {
  int n,k,c;
  cin>>n>>k>>c;
  
  string s;
  cin>>s;
  char p[n];
  
  for(int i=0; i<n; i++) p[i] = s.at(i);
  
  int a[k];
  int b[k];
  
  int count = 0;
  
  for(int i=1; i<=n;){
    if(p[i-1]=='o'){
      a[count]=i;
      count++;
      i = i+c+1;
    }
    else i++;
  }
  
  count = 1;
  
  for(int i=n; i>=1;){
    if(p[i-1]=='o'){
      b[k-count]=i;
      count++;
      i = i-c-1;
    }
    else i--;
  }
  
  for(int i=0; i<k; i++){
    if(a[i]==b[i]) cout << b[i] << endl;
  }
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
  char p[n];
  
  for(int i=0; i<n; i++) p[i] = s.at(i);
  
  int a[k];
  int b[k];
  
  int count = 0;
  
  for(int i=1; i<=n;){
    if(p[i-1]=='o'){
      a[count]=i;
      count++;
      i = i+c+1;
    }
    else i++;
  }
  
  count = 1;
  
  for(int i=n; i>=1;){
    if(p[i-1]=='o'){
      b[k-count]=i;
      count++;
      i = i-c-1;
    }
    else i--;
  }
  
  for(int i=0; i<k; i++){
    if(a[i]==b[i]) cout << b[i] << endl;
  }
}
```