/*
@author: sharrad99
*/
#include<bits/stdc++.h>
using namespace std;
int main(){

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;

	vector<int> a(n, 0), b(n, 0);
	int cnt = 1;
	for(int i = 0; i < n; i++) {
		if(s[i] == 'o') {
			a[i] = cnt++;
			i += c;
		}

		if(cnt > k) break;
	}

	cnt = k;
	for(int i = n - 1; i >= 0; i--) {
		if(s[i] == 'o') {
			b[i] = cnt--;
			i -= c;
		}

		if(cnt < 1) break;
	}

	for(int i = 0; i < n; i++) {
		if(a[i] and a[i] == b[i]) {
			cout << (i + 1) << '\n';
		}
	}
}
### 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
/*
@author: sharrad99
*/
#include<bits/stdc++.h>
using namespace std;
int main(){

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;

	vector<int> a(n, 0), b(n, 0);
	int cnt = 1;
	for(int i = 0; i < n; i++) {
		if(s[i] == 'o') {
			a[i] = cnt++;
			i += c;
		}

		if(cnt > k) break;
	}

	cnt = k;
	for(int i = n - 1; i >= 0; i--) {
		if(s[i] == 'o') {
			b[i] = cnt--;
			i -= c;
		}

		if(cnt < 1) break;
	}

	for(int i = 0; i < n; i++) {
		if(a[i] and a[i] == b[i]) {
			cout << (i + 1) << '\n';
		}
	}
}
```