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

int main(){
  int n, k, c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vector<int> L(k), R(k);
  int i=0;
  int num = 0;
  while (i < n) {
    if (s[i] == 'o') {
      if (num > k) return 0;
      L[num++] = i + 1;
      i += c;
    }
    i++;
  }
  i = n-1;
  num = k - 1;
  while (i >= 0) {
    if (s[i] == 'o') {
      if (num < 0) return 0;
      R[num--] = i + 1;
      i -= c;
    }

    i--;
  }

  for (int j=0; j<k; j++) {
    if (R[j] == L[j]) cout << R[j] << 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;
using ll = long long;

int main(){
  int n, k, c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vector<int> L(k), R(k);
  int i=0;
  int num = 0;
  while (i < n) {
    if (s[i] == 'o') {
      if (num > k) return 0;
      L[num++] = i + 1;
      i += c;
    }
    i++;
  }
  i = n-1;
  num = k - 1;
  while (i >= 0) {
    if (s[i] == 'o') {
      if (num < 0) return 0;
      R[num--] = i + 1;
      i -= c;
    }

    i--;
  }

  for (int j=0; j<k; j++) {
    if (R[j] == L[j]) cout << R[j] << endl;
  }

  return 0;
}
```