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

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  string s;
  int n, k, c;
  cin >> n >> k >> c >> s;

  vector<int> l(k), r(k);
  for (int i = 0, j = 0; i < n && j < k; ++i) {
    if (s[i] == 'x')  continue;
    l[j++] = i;
    i += c;
  }

  for (int i = n-1, j = k-1; i >= 0 && j >= 0; --i) {
    if (s[i] == 'x')  continue;
    r[j--] = i;
    i -= c;
  }

  int ans = 0;
  for (int i = 0; i < k; ++i)
    if (l[i] == r[i]) 
      cout << l[i] + 1 << '\n';
}


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

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  string s;
  int n, k, c;
  cin >> n >> k >> c >> s;

  vector<int> l(k), r(k);
  for (int i = 0, j = 0; i < n && j < k; ++i) {
    if (s[i] == 'x')  continue;
    l[j++] = i;
    i += c;
  }

  for (int i = n-1, j = k-1; i >= 0 && j >= 0; --i) {
    if (s[i] == 'x')  continue;
    r[j--] = i;
    i -= c;
  }

  int ans = 0;
  for (int i = 0; i < k; ++i)
    if (l[i] == r[i]) 
      cout << l[i] + 1 << '\n';
}


```