#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
string s;int cnt;
int n,m,k,a[N],b[N];
int main()
{
	cin>>n>>m>>k>>s;
	int last=-1e8;cnt=0;
	for(int i=0;i<n;i++)
	{
		if(i-last-1>=k&&s[i]=='o')
		{
			cnt++;
			a[cnt]=i;
			last=i;
		}
		if(cnt==m)break;
	}
	last=1e8;cnt=m;
	for(int i=n-1;i>=0;i--)
	{
		if(last-i-1>=k&&s[i]=='o')
		{
			b[cnt--]=i;
			last=i;
		}
		if(!cnt)break;
	}
	for(int i=1;i<=m;i++)
	{
		if(a[i]==b[i])
		{
			cout<<a[i]+1<<"\n";
		}
	}
	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;
const int N=2e5+5;
string s;int cnt;
int n,m,k,a[N],b[N];
int main()
{
	cin>>n>>m>>k>>s;
	int last=-1e8;cnt=0;
	for(int i=0;i<n;i++)
	{
		if(i-last-1>=k&&s[i]=='o')
		{
			cnt++;
			a[cnt]=i;
			last=i;
		}
		if(cnt==m)break;
	}
	last=1e8;cnt=m;
	for(int i=n-1;i>=0;i--)
	{
		if(last-i-1>=k&&s[i]=='o')
		{
			b[cnt--]=i;
			last=i;
		}
		if(!cnt)break;
	}
	for(int i=1;i<=m;i++)
	{
		if(a[i]==b[i])
		{
			cout<<a[i]+1<<"\n";
		}
	}
	return 0;
}
```