#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define endl "\n"

int main()
{
  i64 N, K, C;
  string S;
  cin >> N >> K >> C >> S;
  vector<i64> w;
  for (i64 i = N - 1; 0 <= i; i--)
    if (w.size() < K && S[i] == 'o')
    {
      w.push_back(i);
      i -= C;
    }
  i64 cnt = 0;
  for (i64 i = 0; i < N; i++)
    if (cnt < K && S[i] == 'o')
    {
      if (w[K - cnt - 1] == i)
        cout << i + 1 << endl;
      cnt++;
      i += C;
    }
  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 i64 = long long;
#define endl "\n"

int main()
{
  i64 N, K, C;
  string S;
  cin >> N >> K >> C >> S;
  vector<i64> w;
  for (i64 i = N - 1; 0 <= i; i--)
    if (w.size() < K && S[i] == 'o')
    {
      w.push_back(i);
      i -= C;
    }
  i64 cnt = 0;
  for (i64 i = 0; i < N; i++)
    if (cnt < K && S[i] == 'o')
    {
      if (w[K - cnt - 1] == i)
        cout << i + 1 << endl;
      cnt++;
      i += C;
    }
  return 0;
}

```