#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main() {
	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	
	vector<int> L, R;
	int inserted = 0;
	for(int i = 0; i < n && L.size() != k; i++) {
		if(s.at(i) == 'o') {
			L.push_back(i);
			i += c;
		}
	}
	
	inserted = 0;
	for(int i = n - 1; i >= 0 && R.size() != k; i--) {
		if(s.at(i) == 'o') {
			R.push_back(i);
			i -= c;
		}
	}
	
	for(int i = 0; i < k; i++) {
		if(L[i] == R[k - i - 1]) cout << L[i] + 1 << endl;
	}
}
### Prompt

Your task is to create 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 <iostream>
#include <vector>
#include <string>

using namespace std;

int main() {
	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	
	vector<int> L, R;
	int inserted = 0;
	for(int i = 0; i < n && L.size() != k; i++) {
		if(s.at(i) == 'o') {
			L.push_back(i);
			i += c;
		}
	}
	
	inserted = 0;
	for(int i = n - 1; i >= 0 && R.size() != k; i--) {
		if(s.at(i) == 'o') {
			R.push_back(i);
			i -= c;
		}
	}
	
	for(int i = 0; i < k; i++) {
		if(L[i] == R[k - i - 1]) cout << L[i] + 1 << endl;
	}
}
```