#include<bits/stdc++.h>
using namespace std; 

const int N=2e5+10;

int n,k,c;
int L[N],R[N];
char str[N];

int main()
{
	cin>>n>>k>>c;
	cin>>(str+1);
	
	int cnt=0;
	for(int i=1,j=0;i<=n;i++) {
		if(str[i]!='x') {
			L[++j]=i;
			i+=c;
			cnt=max(cnt,j);
		}
	}
	if(cnt>k) return 0;
	for(int i=n,j=cnt;i>=1;i--) {
		if(str[i]!='x') {
			R[j--]=i;
			i-=c;
		}
	}
	
	for(int i=1;i<=cnt;i++) {
		//cout<<L[i]<<' '<<R[i]<<'\n';
		if(L[i]==R[i]) cout<<L[i]<<'\n';
	}
	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; 

const int N=2e5+10;

int n,k,c;
int L[N],R[N];
char str[N];

int main()
{
	cin>>n>>k>>c;
	cin>>(str+1);
	
	int cnt=0;
	for(int i=1,j=0;i<=n;i++) {
		if(str[i]!='x') {
			L[++j]=i;
			i+=c;
			cnt=max(cnt,j);
		}
	}
	if(cnt>k) return 0;
	for(int i=n,j=cnt;i>=1;i--) {
		if(str[i]!='x') {
			R[j--]=i;
			i-=c;
		}
	}
	
	for(int i=1;i<=cnt;i++) {
		//cout<<L[i]<<' '<<R[i]<<'\n';
		if(L[i]==R[i]) cout<<L[i]<<'\n';
	}
	return 0;
}
```