#include <bits/stdc++.h>
using namespace std;
int main()
{
	int n,k,c;
	string s;
	cin>>n>>k>>c;
	cin>>s;
	vector<int> v1,v2;
	int last=-1e9,kk=k;
	for(int i=0;i<s.size();i++)
	{
		
		if(s[i]=='o'&&i>last+c){
			v1.push_back(i);
			kk--;
			last=i;
		}
		if(kk==0)break;
	}
	int up=-1e9;
	kk=k;
	reverse(s.begin(),s.end());
	for(int i=0;i<s.size();i++)
	{
		if(s[i]=='o'&&i>up+c){
			v2.push_back(s.size()-i-1);
			kk--;
			up=i;
		}
		if(kk==0)break;
	}
	reverse(v2.begin(),v2.end());
	for(int i=0;i<k;i++)
	{
		if(v1[i]==v2[i])cout<<v1[i]+1<<endl;
	}
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
int main()
{
	int n,k,c;
	string s;
	cin>>n>>k>>c;
	cin>>s;
	vector<int> v1,v2;
	int last=-1e9,kk=k;
	for(int i=0;i<s.size();i++)
	{
		
		if(s[i]=='o'&&i>last+c){
			v1.push_back(i);
			kk--;
			last=i;
		}
		if(kk==0)break;
	}
	int up=-1e9;
	kk=k;
	reverse(s.begin(),s.end());
	for(int i=0;i<s.size();i++)
	{
		if(s[i]=='o'&&i>up+c){
			v2.push_back(s.size()-i-1);
			kk--;
			up=i;
		}
		if(kk==0)break;
	}
	reverse(v2.begin(),v2.end());
	for(int i=0;i<k;i++)
	{
		if(v1[i]==v2[i])cout<<v1[i]+1<<endl;
	}
	return 0;
}
```