/**
 *    author:  math2do
 *    created: 07.04.2020 10:01:16
**/

#include <bits/stdc++.h>

using namespace std;

int main() {
  int n, k, c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vector <int> rankR(n, 0);
  int rank = k;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == 'o') {
      rankR[i] = rank--;
      i -= c;
    }
  }
  rank = 1;
  for (int i = 0; i < n; ++i) {
    if (s[i] == 'o') {
      if (rankR[i] == rank++) {
        cout << (i + 1) << '\n';
      }
      i += c;
    }
  }
  return 0;
}
### Prompt

Create a solution in Cpp for 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
/**
 *    author:  math2do
 *    created: 07.04.2020 10:01:16
**/

#include <bits/stdc++.h>

using namespace std;

int main() {
  int n, k, c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vector <int> rankR(n, 0);
  int rank = k;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == 'o') {
      rankR[i] = rank--;
      i -= c;
    }
  }
  rank = 1;
  for (int i = 0; i < n; ++i) {
    if (s[i] == 'o') {
      if (rankR[i] == rank++) {
        cout << (i + 1) << '\n';
      }
      i += c;
    }
  }
  return 0;
}
```