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

signed main() {
  int n,c,k;
  string s;
  cin >> n >> k >> c >> s;

  vector<int> l(k+1,-1), r(k+1,-1);
  int cnt = 1;
  for (int i = 0; i < n; i++) {
    if(s[i] == 'o') {
      l[cnt] = i;
      i+=c;
      cnt++;
    }
    if(cnt > k) break;
  }
  cnt = k;
  for (int i = n - 1; i >= 0; i--) {
    if(s[i] == 'o') {
      r[cnt] = i;
      i-=c;
      cnt--;
    }
    if(cnt < 0) break;
  }

  for (int i = 1; i <= k; i++) {
    if(l[i] == r[i]) cout << l[i] + 1 << '\n';
  }


}
### Prompt

Please provide a cpp coded solution 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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;

signed main() {
  int n,c,k;
  string s;
  cin >> n >> k >> c >> s;

  vector<int> l(k+1,-1), r(k+1,-1);
  int cnt = 1;
  for (int i = 0; i < n; i++) {
    if(s[i] == 'o') {
      l[cnt] = i;
      i+=c;
      cnt++;
    }
    if(cnt > k) break;
  }
  cnt = k;
  for (int i = n - 1; i >= 0; i--) {
    if(s[i] == 'o') {
      r[cnt] = i;
      i-=c;
      cnt--;
    }
    if(cnt < 0) break;
  }

  for (int i = 1; i <= k; i++) {
    if(l[i] == r[i]) cout << l[i] + 1 << '\n';
  }


}
```