#include<bits/stdc++.h>

using namespace std;

int main(){
  int N, K, C;
  cin >> N >> K >> C;
  
  string S;
  cin >> S;
  
  vector<int> left(K, 0);
  vector<int> right(K, 0);
  
  for(int k = 0, i = 0; k < K; k++){
    while(i < N && S[i] == 'x') i++;
    left[k] = i;
    i += C + 1;
    if(i > N) i = N;
  }
  
  for(int k = K - 1, i = N - 1; k >= 0; k--){
    while(i >= 0 && S[i] == 'x') i--;
    right[k] = i;
    i -= C + 1;
    if(i < -1) i = -1;
  }
  
  for(int i = 0; i < K; i++){
    if(left[i] == right[i]) cout << left[i] + 1 << endl;
  }
  
  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(){
  int N, K, C;
  cin >> N >> K >> C;
  
  string S;
  cin >> S;
  
  vector<int> left(K, 0);
  vector<int> right(K, 0);
  
  for(int k = 0, i = 0; k < K; k++){
    while(i < N && S[i] == 'x') i++;
    left[k] = i;
    i += C + 1;
    if(i > N) i = N;
  }
  
  for(int k = K - 1, i = N - 1; k >= 0; k--){
    while(i >= 0 && S[i] == 'x') i--;
    right[k] = i;
    i -= C + 1;
    if(i < -1) i = -1;
  }
  
  for(int i = 0; i < K; i++){
    if(left[i] == right[i]) cout << left[i] + 1 << endl;
  }
  
  return 0;
}
```