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

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

  vector<int> l(k, 0), r(k, 0);
  int cur = 0, d = 1e9;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'o' and d >= c and cur < k) {
      l[cur++] = i;
      d = 0;
    } else {
      ++d;
    }
  }
  cur = k-1, d = 1e9;
  for (int i = n-1; i >= 0; i--) {
    if (s[i] == 'o' and d >= c and cur >= 0) {
      r[cur--] = i;
      d = 0;
    } else {
      ++d;
    }
  }

  for (int i = 0; i < k; i++) {
    if (l[i] == r[i]) {
      cout << l[i]+1 << endl;
    }
  }

  return 0;
}
### 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>
using namespace std;

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

  vector<int> l(k, 0), r(k, 0);
  int cur = 0, d = 1e9;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'o' and d >= c and cur < k) {
      l[cur++] = i;
      d = 0;
    } else {
      ++d;
    }
  }
  cur = k-1, d = 1e9;
  for (int i = n-1; i >= 0; i--) {
    if (s[i] == 'o' and d >= c and cur >= 0) {
      r[cur--] = i;
      d = 0;
    } else {
      ++d;
    }
  }

  for (int i = 0; i < k; i++) {
    if (l[i] == r[i]) {
      cout << l[i]+1 << endl;
    }
  }

  return 0;
}
```