#include <bits/stdc++.h>

#define REP(i, n) for (int i = 0; i < (int)(n); ++i)

using namespace std;

int main() {
  int N, K, C;
  cin >> N >> K >> C;
  string S;
  cin >> S;

  vector<int> earliest;
  int d = 0;
  REP(i, K) {
    while (S[d] == 'x') ++d;
    // cout << i << " " << d << endl;
    earliest.push_back(d);
    d += C + 1;
  }

  vector<int> latest;
  d = N - 1;
  REP(i, K) {
    while (S[d] == 'x') --d;
    // cout << i << " " << d << endl;
    latest.push_back(d);
    d -= C + 1;
  }

  REP(i, K) if (earliest[i] == latest[latest.size() - 1 - i]) {
    cout << earliest[i] + 1 << endl;
  }
}

### Prompt

Create a solution in Cpp for 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>

#define REP(i, n) for (int i = 0; i < (int)(n); ++i)

using namespace std;

int main() {
  int N, K, C;
  cin >> N >> K >> C;
  string S;
  cin >> S;

  vector<int> earliest;
  int d = 0;
  REP(i, K) {
    while (S[d] == 'x') ++d;
    // cout << i << " " << d << endl;
    earliest.push_back(d);
    d += C + 1;
  }

  vector<int> latest;
  d = N - 1;
  REP(i, K) {
    while (S[d] == 'x') --d;
    // cout << i << " " << d << endl;
    latest.push_back(d);
    d -= C + 1;
  }

  REP(i, K) if (earliest[i] == latest[latest.size() - 1 - i]) {
    cout << earliest[i] + 1 << endl;
  }
}

```