#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;

int n,k,c,a[N],b[N];
string s;
 
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k>>c>>s; 
	int cnt=0;
	for(int i=0;i<s.size()&&cnt<k;i++){
		if(s[i]=='o'){
			cnt++;
			a[i]=cnt;
			i+=c;
		}
	}
	cnt=0;
	for(int i=s.size()-1;i>=0&&cnt<k;i--){
		if(s[i]=='o'){
			cnt++;
			b[i]=cnt;
			i-=c;
		}
	}
	for(int i=0;i<s.size();i++){
		if(a[i]!=0&&b[i]!=0&&a[i]+b[i]==k+1) cout<<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;

int n,k,c,a[N],b[N];
string s;
 
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k>>c>>s; 
	int cnt=0;
	for(int i=0;i<s.size()&&cnt<k;i++){
		if(s[i]=='o'){
			cnt++;
			a[i]=cnt;
			i+=c;
		}
	}
	cnt=0;
	for(int i=s.size()-1;i>=0&&cnt<k;i--){
		if(s[i]=='o'){
			cnt++;
			b[i]=cnt;
			i-=c;
		}
	}
	for(int i=0;i<s.size();i++){
		if(a[i]!=0&&b[i]!=0&&a[i]+b[i]==k+1) cout<<i+1<<'\n';
	}
	return 0;
}

```