#include <iostream>
#include <vector>
using namespace std;

int main() {
	int N, K, C;
	cin >> N >> K >> C;
	string S;
	cin >> S;
	vector<int> Left, Right;
	int i = 0;
	int num = 0;
	while (i<N) {
		if (S[i] == 'o') {
			Left.push_back(i);
			i += C + 1;
			num++;
		}
		else i++;
		if (num == K) break;
	}
	i = N - 1;
	num = 0;
	while (i >= 0) {
		if (S[i] == 'o') {
			Right.push_back(i);
			i -= (C + 1);
			num++;
		}
		else i--;
		if (num == K) break;
	}
	for (int i = 0; i < K; i++) {
		if (Left[i] == Right[K-i-1] ) cout << Left[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>
using namespace std;

int main() {
	int N, K, C;
	cin >> N >> K >> C;
	string S;
	cin >> S;
	vector<int> Left, Right;
	int i = 0;
	int num = 0;
	while (i<N) {
		if (S[i] == 'o') {
			Left.push_back(i);
			i += C + 1;
			num++;
		}
		else i++;
		if (num == K) break;
	}
	i = N - 1;
	num = 0;
	while (i >= 0) {
		if (S[i] == 'o') {
			Right.push_back(i);
			i -= (C + 1);
			num++;
		}
		else i--;
		if (num == K) break;
	}
	for (int i = 0; i < K; i++) {
		if (Left[i] == Right[K-i-1] ) cout << Left[i] + 1 << endl;
	}
}
```