#include<iostream>


using namespace std;


int main() {
	long n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	int* front = new int[k+2];
	int* back = new int[k+2];
	front[0] = back[0] = -c-1;
	front[k + 1] = back[k + 1] = n +c;
	for (int i = 1;i <= k;i++) {
		for (int j = front[i - 1] + c+1;;j++) {
			if (s[j] == 'o') {
				front[i] = j;
				break;
			}
		}
		for (int j = back[k - i + 2] - c-1;;j--) {
			if (s[j] == 'o') {
				back[k-i+1] = j;
				break;
			}
		}
	}
	for (int i = 1;i <= k;i++) {
		if (front[i] == back[i]) {
			cout << front[i]+1 << "\n";
		}
	}


}

### Prompt

Create a solution in cpp for 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>


using namespace std;


int main() {
	long n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	int* front = new int[k+2];
	int* back = new int[k+2];
	front[0] = back[0] = -c-1;
	front[k + 1] = back[k + 1] = n +c;
	for (int i = 1;i <= k;i++) {
		for (int j = front[i - 1] + c+1;;j++) {
			if (s[j] == 'o') {
				front[i] = j;
				break;
			}
		}
		for (int j = back[k - i + 2] - c-1;;j--) {
			if (s[j] == 'o') {
				back[k-i+1] = j;
				break;
			}
		}
	}
	for (int i = 1;i <= k;i++) {
		if (front[i] == back[i]) {
			cout << front[i]+1 << "\n";
		}
	}


}

```