#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); i++)
using ll = long long;

int main(){
  int n, k, c; cin >> n >> k >> c;
  string s; cin >> s;
  
  auto getPosition = [&](){
    vector<int> res;
  for(int i=0; i<n && res.size()<=k ;){
    if(s[i]=='x') i++;
    else{ res.push_back(i); i+= c+1;}
  }    
    return res;
  };
  
  vector<int> l, r;
  l = getPosition();
  reverse(s.begin(), s.end());
  r = getPosition();
  rep(i, r.size()) r[i] =n-1-r[i];
  reverse(r.begin(), r.end());

  rep(i, k){
    if(l[i] == r[i]) cout << l[i]+1 << endl;
  }
  
  return 0;
}
### 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;
#define rep(i, n) for(int i=0; i<(n); i++)
using ll = long long;

int main(){
  int n, k, c; cin >> n >> k >> c;
  string s; cin >> s;
  
  auto getPosition = [&](){
    vector<int> res;
  for(int i=0; i<n && res.size()<=k ;){
    if(s[i]=='x') i++;
    else{ res.push_back(i); i+= c+1;}
  }    
    return res;
  };
  
  vector<int> l, r;
  l = getPosition();
  reverse(s.begin(), s.end());
  r = getPosition();
  rep(i, r.size()) r[i] =n-1-r[i];
  reverse(r.begin(), r.end());

  rep(i, k){
    if(l[i] == r[i]) cout << l[i]+1 << endl;
  }
  
  return 0;
}
```