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

Construct a CPP code solution to the problem outlined:

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;
vector<int> f,b;
int n,k,c;
string s;
int main()
{
	cin>>n>>k>>c>>s;
	int last=-1e9;
	for(int i=1;i<=n;i++)
	{
		if(s[i-1]=='o'&&last+c<i)
		{
			last=i;
			f.push_back(i);
		}
		if(f.size()==k)
			break;
	}
	int stock=1e9;
	for(int i=n;i>=1;i--)
	{
		if(s[i-1]=='o'&&stock>=c)
		{
			b.push_back(i);
			stock=0; 
		}	
		else
			stock++;
		if(b.size()==k)
			break;
	}
	reverse(b.begin(),b.end());
	for(int i=0;i<k;i++)
		if(b[i]==f[i])
			cout<<f[i]<<endl;
	return 0;
}
```