#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 5;

int n, k, c, tot;
char s[N];
int lf[N], rt[N];

int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> c;
  cin >> (s + 1);
  int right = 0; 
  tot = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == 'o' && i >= right) 
      lf[++tot] = i, right = i + c + 1;
  }
  if (tot > k)  
    return 0;
  int left = n; 
  for (int i = n, l = 0; i >= 1; i--)
    if (s[i] == 'o' && i <= left)
      left = i - c - 1, rt[tot - (l++)] = i; 
  for (int i = 1; i <= tot; i++)
    if (lf[i] == rt[i])
      cout << lf[i] << '\n';
  return 0; 
}
### Prompt

Develop a solution in CPP 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;

const int N = 2e5 + 5;

int n, k, c, tot;
char s[N];
int lf[N], rt[N];

int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> c;
  cin >> (s + 1);
  int right = 0; 
  tot = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == 'o' && i >= right) 
      lf[++tot] = i, right = i + c + 1;
  }
  if (tot > k)  
    return 0;
  int left = n; 
  for (int i = n, l = 0; i >= 1; i--)
    if (s[i] == 'o' && i <= left)
      left = i - c - 1, rt[tot - (l++)] = i; 
  for (int i = 1; i <= tot; i++)
    if (lf[i] == rt[i])
      cout << lf[i] << '\n';
  return 0; 
}
```