#include <bits/stdc++.h>
using namespace std;
int n,k,c,cnt;
string s;
set<int> se;

signed main(){
  cin>>n>>k>>c>>s;
  if(!c){
    for(char p:s)k-=p=='o';
    if(!k)for(int i=0;i<n;i++)if(s[i]=='o')cout<<i+1<<" ";
    cout<<endl;
    return 0;
  }
  for(int i=n-1;cnt<k;i--){
    if(s[i]=='o'){
      se.insert(i);
      cnt++;
      i-=c;
    }
  }
  cnt=0;
  for(int i=0;cnt<k;i++){
    if(s[i]=='o'){
      if(se.count(i))cout<<i+1<<" ";
      cnt++;
      i+=c;
    }
  }
  cout<<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 n,k,c,cnt;
string s;
set<int> se;

signed main(){
  cin>>n>>k>>c>>s;
  if(!c){
    for(char p:s)k-=p=='o';
    if(!k)for(int i=0;i<n;i++)if(s[i]=='o')cout<<i+1<<" ";
    cout<<endl;
    return 0;
  }
  for(int i=n-1;cnt<k;i--){
    if(s[i]=='o'){
      se.insert(i);
      cnt++;
      i-=c;
    }
  }
  cnt=0;
  for(int i=0;cnt<k;i++){
    if(s[i]=='o'){
      if(se.count(i))cout<<i+1<<" ";
      cnt++;
      i+=c;
    }
  }
  cout<<endl;
}

```