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

int main(){
  int n, k, c;
  string s;
  cin >> n >> k >> c >> s;
  vector<int> work(n,-1);
  int cnt = 0;
  for(int i=0; i<n; ++i){
    if(s[i] == 'o'){
      ++cnt;
      work[i] = cnt;
      if(cnt == k) break;
      i += c;
    }
  }
  vector<int> ans(0);
  for(int i=n-1; i>=0; --i){
    if(s[i] == 'o'){
      if(work[i] == cnt) ans.push_back(i+1);
      --cnt;
      if(cnt == 0) break;
      i -= c;
    }
  }
  reverse(ans.begin(),ans.end());
  for(int a: ans){
    printf("%d\n",a);
  }
}
### Prompt

In Cpp, your task is to solve 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;
  string s;
  cin >> n >> k >> c >> s;
  vector<int> work(n,-1);
  int cnt = 0;
  for(int i=0; i<n; ++i){
    if(s[i] == 'o'){
      ++cnt;
      work[i] = cnt;
      if(cnt == k) break;
      i += c;
    }
  }
  vector<int> ans(0);
  for(int i=n-1; i>=0; --i){
    if(s[i] == 'o'){
      if(work[i] == cnt) ans.push_back(i+1);
      --cnt;
      if(cnt == 0) break;
      i -= c;
    }
  }
  reverse(ans.begin(),ans.end());
  for(int a: ans){
    printf("%d\n",a);
  }
}
```