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

int n, k, c;
string s;

vector<int> greed(){
  vector<int> res(n, -10);
  
  int cnt = 1;
  for(int i = 0; i < n || cnt <= k; ++i){
    if(s.at(i) == 'x') continue;
    res.at(i) = cnt++;
    i += c;
  }
  
  return res;
}

int main(){
  cin >> n >> k >> c >> s;
  
  vector<int> l, r;
  l = greed();
  reverse(s.begin(), s.end());
  r = greed();
  reverse(r.begin(), r.end());
  
  for(int i = 0; i < n; ++i){
    if(l.at(i) < 0) continue;
    if(l.at(i) == k - r.at(i) + 1) cout << i + 1 << 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 n, k, c;
string s;

vector<int> greed(){
  vector<int> res(n, -10);
  
  int cnt = 1;
  for(int i = 0; i < n || cnt <= k; ++i){
    if(s.at(i) == 'x') continue;
    res.at(i) = cnt++;
    i += c;
  }
  
  return res;
}

int main(){
  cin >> n >> k >> c >> s;
  
  vector<int> l, r;
  l = greed();
  reverse(s.begin(), s.end());
  r = greed();
  reverse(r.begin(), r.end());
  
  for(int i = 0; i < n; ++i){
    if(l.at(i) < 0) continue;
    if(l.at(i) == k - r.at(i) + 1) cout << i + 1 << endl;
  }
}
```