#include <bits/stdc++.h>
using namespace std;
const int inf=1000000;
int n,k,c;
string s;
int main(){
	cin>>n>>k>>c>>s;
	vector<int> g1(k),g2(k);
	{
		int lst=-inf;
		int p=0;
		for(int i=0;i<n&&p<k;++i){
			if(s[i]=='o'&&i>lst+c){
				g1[p++]=i;
				lst=i;
			}
		}
	}
	{
		int lst=inf;
		int p=k-1;
		for(int i=n-1;i>=0&&p>=0;--i){
			if(s[i]=='o'&&i<lst-c){
				g2[p--]=i;
				lst=i;
			}
		}
	}
	for(int i=0;i<k;++i){
		if(g1[i]==g2[i]){
			cout<<g1[i]+1<<'\n';
		}
	}
	return 0;
} 
### Prompt

Generate a CPP solution to 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;
const int inf=1000000;
int n,k,c;
string s;
int main(){
	cin>>n>>k>>c>>s;
	vector<int> g1(k),g2(k);
	{
		int lst=-inf;
		int p=0;
		for(int i=0;i<n&&p<k;++i){
			if(s[i]=='o'&&i>lst+c){
				g1[p++]=i;
				lst=i;
			}
		}
	}
	{
		int lst=inf;
		int p=k-1;
		for(int i=n-1;i>=0&&p>=0;--i){
			if(s[i]=='o'&&i<lst-c){
				g2[p--]=i;
				lst=i;
			}
		}
	}
	for(int i=0;i<k;++i){
		if(g1[i]==g2[i]){
			cout<<g1[i]+1<<'\n';
		}
	}
	return 0;
} 
```