#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
  int n, k, c;
  scanf("%d %d %d", &n, &k, &c);
  string s;
  cin >> s;

  vector<bool> l(n), r(n);
  int m = 0, lp = 0, rp = n;
  for (int i = 0; i < n; i++)
    if (i >= lp && s[i] == 'o') {
      l[i] = true;
      lp = i + c + 1;
      m++;
    }
  for (int i = n - 1; i >= 0; i--)
    if (i < rp && s[i] == 'o') {
      r[i] = true;
      rp = i - c;
    }

  if (m == k) {
    for (int i = 0; i < n; i++)
      if (l[i] && r[i])
        printf("%d\n", i + 1);
  }

  return 0;
}
### Prompt

Please formulate a Cpp solution to 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 <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
  int n, k, c;
  scanf("%d %d %d", &n, &k, &c);
  string s;
  cin >> s;

  vector<bool> l(n), r(n);
  int m = 0, lp = 0, rp = n;
  for (int i = 0; i < n; i++)
    if (i >= lp && s[i] == 'o') {
      l[i] = true;
      lp = i + c + 1;
      m++;
    }
  for (int i = n - 1; i >= 0; i--)
    if (i < rp && s[i] == 'o') {
      r[i] = true;
      rp = i - c;
    }

  if (m == k) {
    for (int i = 0; i < n; i++)
      if (l[i] && r[i])
        printf("%d\n", i + 1);
  }

  return 0;
}
```