/*input
16 4 3
ooxxoxoxxxoxoxxo
*/

#include <bits/stdc++.h>
#define ll long long

using namespace std;

void run() {
	int n, k, c, i;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> first, last;
	for(i=n-1; i>=0; i--) {
		if(s[i]=='o') {
			last.emplace_back(i);
			i -= c;
		}
		if(last.size() == k) 
			break;
	}
	for(i=0; i<n; i++) {
		if(s[i]=='o') {
			first.emplace_back(i);
			i += c;
		}
		if(first.size() == k) 
			break;
	}
	for(i=0; i<k; i++)
		if(first[i] == last[k - 1 - i])
			cout << first[i] + 1 << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);		
	run();
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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
/*input
16 4 3
ooxxoxoxxxoxoxxo
*/

#include <bits/stdc++.h>
#define ll long long

using namespace std;

void run() {
	int n, k, c, i;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> first, last;
	for(i=n-1; i>=0; i--) {
		if(s[i]=='o') {
			last.emplace_back(i);
			i -= c;
		}
		if(last.size() == k) 
			break;
	}
	for(i=0; i<n; i++) {
		if(s[i]=='o') {
			first.emplace_back(i);
			i += c;
		}
		if(first.size() == k) 
			break;
	}
	for(i=0; i<k; i++)
		if(first[i] == last[k - 1 - i])
			cout << first[i] + 1 << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);		
	run();
}

```