#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n, k, c;
  string s;
  cin >> n >> k >> c >> s;
  vector < int > l, r;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'o' && l.size() < k) {
      l.push_back(i);
      i += c;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == 'o' && r.size() < k) {
      r.push_back(i);
      i -= c;
    }
  }
  reverse(r.begin(), r.end());
  for (int i = 0; i < l.size() && i < r.size(); i++) {
    if (l[i] == r[i]) cout << l[i] + 1 << endl;
  }
}

### 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;

typedef long long int ll;

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n, k, c;
  string s;
  cin >> n >> k >> c >> s;
  vector < int > l, r;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'o' && l.size() < k) {
      l.push_back(i);
      i += c;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == 'o' && r.size() < k) {
      r.push_back(i);
      i -= c;
    }
  }
  reverse(r.begin(), r.end());
  for (int i = 0; i < l.size() && i < r.size(); i++) {
    if (l[i] == r[i]) cout << l[i] + 1 << endl;
  }
}

```