#include <bits/stdc++.h>

#define pb push_back

using namespace std;

int main(){
	int n, k, c;
	vector<int> early, late;
	string s;
	cin>>n>>k>>c;
	//getline(cin, s);
	cin>>s;
	//early-greedy
	int idx = 0;
	while(early.size() < k){
		if(s[idx] == 'o'){
			early.pb(idx);
			idx+=c+1;
		}
		else{
			idx++;
		}
	}

	//late-greedy
	idx = n-1;
	while(late.size() < k){
		if(s[idx] == 'o'){
			late.pb(idx);
			idx-=c+1;
		}
		else{
			idx--;
		}
	}
	reverse(late.begin(), late.end());

	for(int i=0;i<k;i++){
		//cout<<early[i]<<" "<<late[i]<<"\n";
		if(early[i] == late[i])
			cout<<early[i]+1<<"\n";
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>

#define pb push_back

using namespace std;

int main(){
	int n, k, c;
	vector<int> early, late;
	string s;
	cin>>n>>k>>c;
	//getline(cin, s);
	cin>>s;
	//early-greedy
	int idx = 0;
	while(early.size() < k){
		if(s[idx] == 'o'){
			early.pb(idx);
			idx+=c+1;
		}
		else{
			idx++;
		}
	}

	//late-greedy
	idx = n-1;
	while(late.size() < k){
		if(s[idx] == 'o'){
			late.pb(idx);
			idx-=c+1;
		}
		else{
			idx--;
		}
	}
	reverse(late.begin(), late.end());

	for(int i=0;i<k;i++){
		//cout<<early[i]<<" "<<late[i]<<"\n";
		if(early[i] == late[i])
			cout<<early[i]+1<<"\n";
	}
	return 0;
}
```