#include<bits/stdc++.h>
using namespace std;
int n,k,c; 
string s;
void solve(){
	cin>>n>>k>>c>>s;
	s = '#' + s;
	vector<int> earliest(k+1,0),latest(k+1,0);
	int i=1,j=1;
	while(i<s.length()&&j<=k){
		if(s[i]=='o'){
			earliest[j++] = i;
			i+=c+1;
		}else{
			i++;
		}
	}
	i=s.length()-1,j=k;
	while(i>0&&j>=1){
		if(s[i]=='o'){
			latest[j--] = i;
			i-=c+1;
		}else{
			i--;
		}
	}
	for(i=1;i<k+1;i++)if(earliest[i]==latest[i])cout<<earliest[i]<<"\n";
}

int main()
{
  solve();
  return 0;
}
### 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
#include<bits/stdc++.h>
using namespace std;
int n,k,c; 
string s;
void solve(){
	cin>>n>>k>>c>>s;
	s = '#' + s;
	vector<int> earliest(k+1,0),latest(k+1,0);
	int i=1,j=1;
	while(i<s.length()&&j<=k){
		if(s[i]=='o'){
			earliest[j++] = i;
			i+=c+1;
		}else{
			i++;
		}
	}
	i=s.length()-1,j=k;
	while(i>0&&j>=1){
		if(s[i]=='o'){
			latest[j--] = i;
			i-=c+1;
		}else{
			i--;
		}
	}
	for(i=1;i<k+1;i++)if(earliest[i]==latest[i])cout<<earliest[i]<<"\n";
}

int main()
{
  solve();
  return 0;
}
```