#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, K, C;
  cin >> N >> K >> C;
  string S;
  cin >> S;
  vector<int> L(N, 0), R(N, 0);
  int cur = 1;
  for (int i = 0; i < N; i++) {
    if (S[i] == 'o') {
      L[i] = cur++;
      i += C;
    }
    if (cur == K + 1) break;
  }
  cur--;
  for (int i = N - 1; i > -1; i--) {
    if (S[i] == 'o') {
      R[i] = cur--;
      i -= C;
    }
    if (cur == -1) break;
  }
  for (int i = 0; i < N; i++) {
    if (L[i] != 0 && L[i] == R[i]) cout << i + 1 << '\n';
  }
  return 0;
}

### Prompt

Generate 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);
  cin.tie(0);
  int N, K, C;
  cin >> N >> K >> C;
  string S;
  cin >> S;
  vector<int> L(N, 0), R(N, 0);
  int cur = 1;
  for (int i = 0; i < N; i++) {
    if (S[i] == 'o') {
      L[i] = cur++;
      i += C;
    }
    if (cur == K + 1) break;
  }
  cur--;
  for (int i = N - 1; i > -1; i--) {
    if (S[i] == 'o') {
      R[i] = cur--;
      i -= C;
    }
    if (cur == -1) break;
  }
  for (int i = 0; i < N; i++) {
    if (L[i] != 0 && L[i] == R[i]) cout << i + 1 << '\n';
  }
  return 0;
}

```